NCERT Class 10 Maths Solutions: Arithmetic Progressions

Question:

Choose the correct choice in the following and justify:

30th term of the AP: 10, 7, 4, …, is ______.

A. 97

B. 77

C. –77

D. – 87

Concept in a Minute:

Explanation:

The given arithmetic progression (AP) is 10, 7, 4, …
The first term ($a_1$) is 10.
To find the common difference (d), we subtract any term from its succeeding term:
$d = 7 – 10 = -3$
$d = 4 – 7 = -3$
So, the common difference (d) is -3.
We need to find the 30th term ($a_{30}$). Using the formula for the nth term of an AP:
$a_n = a_1 + (n-1)d$
Substitute $n = 30$, $a_1 = 10$, and $d = -3$:
$a_{30} = 10 + (30-1)(-3)$
$a_{30} = 10 + (29)(-3)$
$a_{30} = 10 – 87$
$a_{30} = -77$

The 30th term of the AP is -77.

Therefore, the correct choice is C.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

$\sqrt{2,\sqrt{8,\sqrt{18,\sqrt{32…}$

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we need to calculate the difference between each pair of consecutive terms. If these differences are all equal, then the sequence is an AP.

Explanation:

First, let’s simplify the given terms of the sequence:
The first term is $\sqrt{2}$.
The second term is $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.
The third term is $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$.
The fourth term is $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$.

So the sequence can be rewritten as: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, …$

Now, let’s check for the common difference between consecutive terms:
Difference between the second and first term: $2\sqrt{2} – \sqrt{2} = (2-1)\sqrt{2} = \sqrt{2}$.
Difference between the third and second term: $3\sqrt{2} – 2\sqrt{2} = (3-2)\sqrt{2} = \sqrt{2}$.
Difference between the fourth and third term: $4\sqrt{2} – 3\sqrt{2} = (4-3)\sqrt{2} = \sqrt{2}$.

Since the difference between consecutive terms is constant, which is $\sqrt{2}$, the given sequence is an Arithmetic Progression.

The common difference, $d = \sqrt{2}$.

To find the next three terms, we continue adding the common difference to the last term:
The fifth term = Fourth term + $d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2}$.
In radical form, $5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}$.

The sixth term = Fifth term + $d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2}$.
In radical form, $6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72}$.

The seventh term = Sixth term + $d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2}$.
In radical form, $7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$.

Therefore, the next three terms are $\sqrt{50}, \sqrt{72}, \sqrt{98}$.

Question:

Find the 31^st term of an A.P. whose 11^th term is 38 and the 16^th term is 73.

Concept in a Minute:

The core concept is the properties of an Arithmetic Progression (A.P.). An A.P. is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). The general form of an A.P. is a, a+d, a+2d, …, where ‘a’ is the first term. The n-th term of an A.P. is given by the formula: a_n = a + (n-1)d. To solve this problem, we’ll use this formula to set up a system of two linear equations with two variables (the first term ‘a’ and the common difference ‘d’) based on the given information about two terms of the A.P.

Explanation:

Let the first term of the Arithmetic Progression (A.P.) be ‘a’ and the common difference be ‘d’.
The formula for the n-th term of an A.P. is given by: a_n = a + (n-1)d.

We are given that the 11th term is 38. So, we can write:
a_11 = a + (11-1)d
38 = a + 10d — (Equation 1)

We are also given that the 16th term is 73. So, we can write:
a_16 = a + (16-1)d
73 = a + 15d — (Equation 2)

Now we have a system of two linear equations with two variables ‘a’ and ‘d’:
1) a + 10d = 38
2) a + 15d = 73

To solve for ‘d’, we can subtract Equation 1 from Equation 2:
(a + 15d) – (a + 10d) = 73 – 38
a + 15d – a – 10d = 35
5d = 35
d = 35 / 5
d = 7

Now that we have the common difference ‘d’, we can substitute its value into either Equation 1 or Equation 2 to find the first term ‘a’. Let’s use Equation 1:
a + 10d = 38
a + 10(7) = 38
a + 70 = 38
a = 38 – 70
a = -32

So, the first term of the A.P. is -32 and the common difference is 7.

We need to find the 31st term of the A.P. Using the formula a_n = a + (n-1)d:
a_31 = a + (31-1)d
a_31 = a + 30d
Substitute the values of ‘a’ and ‘d’ that we found:
a_31 = -32 + 30(7)
a_31 = -32 + 210
a_31 = 178

Therefore, the 31st term of the A.P. is 178.

Question:

In the following situation, involved make an arithmetic progression? and why?

The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). To determine if a situation forms an AP, we need to check if the difference between successive quantities remains the same.

Explanation:

Let the initial amount of air in the cylinder be ‘A’.
In the first step, the vacuum pump removes 1/4 of the air, so the amount removed is (1/4)A.
The air remaining in the cylinder after the first step is A – (1/4)A = (3/4)A.

In the second step, the vacuum pump removes 1/4 of the *remaining* air. The remaining air is (3/4)A.
So, the amount removed in the second step is (1/4) * (3/4)A = (3/16)A.
The air remaining in the cylinder after the second step is (3/4)A – (3/16)A = (12/16)A – (3/16)A = (9/16)A.

Let’s look at the sequence of air remaining in the cylinder:
First term (a1) = A
Second term (a2) = (3/4)A
Third term (a3) = (9/16)A

Now let’s check the difference between consecutive terms:
d1 = a2 – a1 = (3/4)A – A = (3/4)A – (4/4)A = -(1/4)A
d2 = a3 – a2 = (9/16)A – (3/4)A = (9/16)A – (12/16)A = -(3/16)A

Since the difference between consecutive terms (d1 and d2) is not constant (-(1/4)A is not equal to -(3/16)A), the situation does not form an arithmetic progression.

Why:
The problem describes a situation where a fixed fraction of the *remaining* quantity is removed each time. This means that the *amount* removed changes with each step, and consequently, the *difference* between consecutive amounts of air remaining is not constant. For an arithmetic progression, the difference between consecutive terms must be constant. In this case, the amount of air remaining forms a geometric progression, not an arithmetic progression.

Question:

Write first four terms of the A.P. when the first term a and the common difference d are given as follows:

$𝑎=−1,𝑑=12$

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’. The first term of the AP is denoted by ‘a’. The formula for the nth term of an AP is given by $a_n = a + (n-1)d$, where $a_n$ is the nth term, a is the first term, n is the term number, and d is the common difference. To find the first four terms, we need to find $a_1$, $a_2$, $a_3$, and $a_4$.

Explanation:

The problem asks us to write the first four terms of an Arithmetic Progression (AP) given the first term ($a$) and the common difference ($d$).
We are given:
First term, $a = -1$
Common difference, $d = 12$

The first term of the AP is simply the given first term, $a_1 = a$.
So, $a_1 = -1$.

The second term ($a_2$) is obtained by adding the common difference ($d$) to the first term ($a_1$).
$a_2 = a_1 + d$
$a_2 = -1 + 12$
$a_2 = 11$.

The third term ($a_3$) is obtained by adding the common difference ($d$) to the second term ($a_2$).
$a_3 = a_2 + d$
$a_3 = 11 + 12$
$a_3 = 23$.

The fourth term ($a_4$) is obtained by adding the common difference ($d$) to the third term ($a_3$).
$a_4 = a_3 + d$
$a_4 = 23 + 12$
$a_4 = 35$.

Alternatively, we can use the formula for the nth term of an AP, $a_n = a + (n-1)d$:
For the first term (n=1):
$a_1 = a + (1-1)d = a + 0d = a$
$a_1 = -1$

For the second term (n=2):
$a_2 = a + (2-1)d = a + 1d = a + d$
$a_2 = -1 + 12$
$a_2 = 11$

For the third term (n=3):
$a_3 = a + (3-1)d = a + 2d$
$a_3 = -1 + 2(12)$
$a_3 = -1 + 24$
$a_3 = 23$

For the fourth term (n=4):
$a_4 = a + (4-1)d = a + 3d$
$a_4 = -1 + 3(12)$
$a_4 = -1 + 36$
$a_4 = 35$

Therefore, the first four terms of the AP are -1, 11, 23, and 35.

Question:

Show that a₁, a_2,…, a_n… form an AP where a_nis defined as below:

a_n = 9 − 5n

Also, find the sum of the first 15 terms.

Concept in a Minute:

To show that a sequence is an Arithmetic Progression (AP), we need to prove that the difference between any two consecutive terms is constant. This constant difference is called the common difference (d). The formula for the nth term of an AP is a_n = a + (n-1)d, where a is the first term. The sum of the first n terms of an AP is given by S_n = n/2 [2a + (n-1)d] or S_n = n/2 (a + l), where l is the last term.

Explanation:

Given the nth term of a sequence as a_n = 9 – 5n.

To show that it forms an AP, we need to find the difference between consecutive terms, i.e., a_n – a_{n-1}.

First term (a_1): Substitute n=1 in the formula.
a_1 = 9 – 5(1) = 9 – 5 = 4.

Second term (a_2): Substitute n=2 in the formula.
a_2 = 9 – 5(2) = 9 – 10 = -1.

Third term (a_3): Substitute n=3 in the formula.
a_3 = 9 – 5(3) = 9 – 15 = -6.

Now let’s find the difference between consecutive terms:
a_2 – a_1 = -1 – 4 = -5.
a_3 – a_2 = -6 – (-1) = -6 + 1 = -5.

Since the difference between consecutive terms is constant (-5), the sequence a_n = 9 – 5n forms an Arithmetic Progression with the first term a = 4 and the common difference d = -5.

Alternatively, we can directly calculate the difference a_n – a_{n-1}:
a_n = 9 – 5n
a_{n-1} = 9 – 5(n-1) = 9 – 5n + 5 = 14 – 5n.
a_n – a_{n-1} = (9 – 5n) – (14 – 5n) = 9 – 5n – 14 + 5n = -5.
Since the difference is a constant (-5), the sequence is an AP.

Now, we need to find the sum of the first 15 terms (S_15).
We have the first term a = 4, the common difference d = -5, and n = 15.
Using the formula S_n = n/2 [2a + (n-1)d]:
S_15 = 15/2 [2(4) + (15-1)(-5)]
S_15 = 15/2 [8 + (14)(-5)]
S_15 = 15/2 [8 – 70]
S_15 = 15/2 [-62]
S_15 = 15 * (-31)
S_15 = -465.

Therefore, the sum of the first 15 terms is -465.

Question:

The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the A.P.

Concept in a Minute:

The question involves an Arithmetic Progression (A.P.). The key concepts are the formula for the nth term of an A.P. (a_n = a + (n-1)d) and how to form and solve linear equations from given information about the terms of an A.P.

Explanation:

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’.
The nth term of an A.P. is given by the formula: a_n = a + (n-1)d

According to the first condition given in the question:
The sum of the 4th and 8th terms of an A.P. is 24.
4th term (a_4) = a + (4-1)d = a + 3d
8th term (a_8) = a + (8-1)d = a + 7d
So, a_4 + a_8 = (a + 3d) + (a + 7d) = 2a + 10d
Given that this sum is 24:
2a + 10d = 24
Dividing the entire equation by 2, we get:
Equation 1: a + 5d = 12

According to the second condition given in the question:
The sum of the 6th and 10th terms is 44.
6th term (a_6) = a + (6-1)d = a + 5d
10th term (a_10) = a + (10-1)d = a + 9d
So, a_6 + a_10 = (a + 5d) + (a + 9d) = 2a + 14d
Given that this sum is 44:
2a + 14d = 44
Dividing the entire equation by 2, we get:
Equation 2: a + 7d = 22

Now we have a system of two linear equations with two variables:
1) a + 5d = 12
2) a + 7d = 22

To solve this system, we can subtract Equation 1 from Equation 2:
(a + 7d) – (a + 5d) = 22 – 12
a + 7d – a – 5d = 10
2d = 10
d = 10 / 2
d = 5

Now substitute the value of ‘d’ (which is 5) into Equation 1:
a + 5d = 12
a + 5(5) = 12
a + 25 = 12
a = 12 – 25
a = -13

So, the first term (a) is -13 and the common difference (d) is 5.

The first three terms of the A.P. are:
First term (a_1) = a = -13
Second term (a_2) = a + d = -13 + 5 = -8
Third term (a_3) = a + 2d = -13 + 2(5) = -13 + 10 = -3

Therefore, the first three terms of the A.P. are -13, -8, and -3.

Question:

Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.

Concept in a Minute:

Arithmetic Progression (AP): A sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d).
Formula for the nth term of an AP: a_n = a + (n-1)d, where a is the first term, n is the term number, and d is the common difference.
Formula for the sum of the first n terms of an AP: S_n = n/2 * [2a + (n-1)d] or S_n = n/2 * [a + a_n], where S_n is the sum of the first n terms, a is the first term, n is the number of terms, d is the common difference, and a_n is the nth term.

Explanation:

The problem asks for the sum of the first 22 terms of an Arithmetic Progression (AP). We are given the common difference (d = 7) and the 22nd term (a_22 = 149). To find the sum of the first 22 terms (S_22), we can use the formula S_n = n/2 * [a + a_n].
First, we need to find the first term (a) of the AP. We can use the formula for the nth term: a_n = a + (n-1)d.
Given: n = 22, d = 7, a_22 = 149.
Substitute these values into the formula:
149 = a + (22-1) * 7
149 = a + 21 * 7
149 = a + 147
Now, solve for ‘a’:
a = 149 – 147
a = 2
So, the first term of the AP is 2.
Now that we have the first term (a), the number of terms (n), and the 22nd term (a_22), we can find the sum of the first 22 terms (S_22) using the formula S_n = n/2 * [a + a_n].
Given: n = 22, a = 2, a_22 = 149.
Substitute these values into the formula:
S_22 = 22/2 * [2 + 149]
S_22 = 11 * [151]
S_22 = 1661

Alternatively, we could use the formula S_n = n/2 * [2a + (n-1)d] after finding the first term ‘a’.
S_22 = 22/2 * [2*2 + (22-1)*7]
S_22 = 11 * [4 + 21*7]
S_22 = 11 * [4 + 147]
S_22 = 11 * [151]
S_22 = 1661

The sum of the first 22 terms of the AP is 1661.

Question:

Show that a₁, a_2,…, a_n… form an AP where a_n is defined as below:

a_n = 3 + 4n

Also, find the sum of the first 15 terms.

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). The n-th term of an AP is given by a_n = a + (n-1)d, where a is the first term and d is the common difference. The sum of the first n terms of an AP is given by S_n = n/2 * [2a + (n-1)d] or S_n = n/2 * (a + l), where l is the last term.

Explanation:

To show that the sequence a_n = 3 + 4n forms an AP, we need to demonstrate that the difference between any two consecutive terms is constant.
Let’s find the (n+1)-th term:
a_{n+1} = 3 + 4(n+1) = 3 + 4n + 4 = 7 + 4n

Now, let’s find the difference between a_{n+1} and a_n:
a_{n+1} – a_n = (7 + 4n) – (3 + 4n)
a_{n+1} – a_n = 7 + 4n – 3 – 4n
a_{n+1} – a_n = 4

Since the difference between consecutive terms (a_{n+1} – a_n) is a constant value of 4, the sequence a_n = 3 + 4n forms an arithmetic progression with a common difference d = 4.

Now, we need to find the sum of the first 15 terms (S_15).
First, let’s find the first term (a_1):
a_1 = 3 + 4(1) = 3 + 4 = 7

We know the common difference d = 4 and n = 15.
Using the formula for the sum of the first n terms of an AP: S_n = n/2 * [2a + (n-1)d]

Substitute the values:
S_15 = 15/2 * [2(7) + (15-1)4]
S_15 = 15/2 * [14 + (14)4]
S_15 = 15/2 * [14 + 56]
S_15 = 15/2 * [70]
S_15 = 15 * 35

To calculate 15 * 35:
15 * 35 = 15 * (30 + 5) = 15 * 30 + 15 * 5 = 450 + 75 = 525

Alternatively, we can find the 15th term (a_15) first:
a_15 = 3 + 4(15) = 3 + 60 = 63
Then use the formula S_n = n/2 * (a + l):
S_15 = 15/2 * (7 + 63)
S_15 = 15/2 * (70)
S_15 = 15 * 35 = 525

Thus, the sum of the first 15 terms is 525.

Question:

Find the 20^th term from the last term of the A.P. 3, 8, 13, …, 253.

Concept in a Minute:

To find a term from the last of an arithmetic progression (A.P.), we can reverse the A.P. and then find the corresponding term from the beginning of the reversed A.P. The formula for the n^th term of an A.P. is $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

Explanation:

The given A.P. is 3, 8, 13, …, 253.
The first term of the A.P., $a = 3$.
The common difference, $d = 8 – 3 = 5$.
The last term of the A.P. is 253.

We need to find the 20^th term from the last term.

Method 1: Reversing the A.P.
If we reverse the A.P., the last term becomes the first term, and the common difference changes its sign.
The reversed A.P. starts with 253.
The new common difference, $d’ = -5$.
We need to find the 20^th term of this reversed A.P.
Using the formula $a_n = a + (n-1)d$:
Here, $a = 253$, $n = 20$, and $d = -5$.
$a_{20} = 253 + (20-1)(-5)$
$a_{20} = 253 + (19)(-5)$
$a_{20} = 253 – 95$
$a_{20} = 158$.

Method 2: Finding the total number of terms first.
First, find the total number of terms in the A.P. using the formula $a_n = a + (n-1)d$, where $a_n$ is the last term.
$253 = 3 + (n-1)5$
$253 – 3 = (n-1)5$
$250 = (n-1)5$
$n-1 = 250 / 5$
$n-1 = 50$
$n = 51$.
So, there are 51 terms in the A.P.

The 20^th term from the last is the same as the $(n – 20 + 1)^{th}$ term from the beginning.
$n – 20 + 1 = 51 – 20 + 1 = 32$.
So, we need to find the 32^nd term of the original A.P.
$a_{32} = a + (32-1)d$
$a_{32} = 3 + (31)5$
$a_{32} = 3 + 155$
$a_{32} = 158$.

Both methods give the same result.
The 20^th term from the last term of the A.P. is 158.

Final Answer: The final answer is $\boxed{158}$

Question:

The first and last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?

Concept in a Minute:

Arithmetic Progression (AP): A sequence of numbers such that the difference between the consecutive terms is constant.
Key Formulas:
1. nth term of an AP: $a_n = a + (n-1)d$
where $a_n$ is the nth term, $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference.
2. Sum of an AP: $S_n = \frac{n}{2}(a + l)$
where $S_n$ is the sum of the first n terms, $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.
Alternatively, $S_n = \frac{n}{2}(2a + (n-1)d)$

Explanation:

The problem provides the first term ($a$), the last term ($l$), and the common difference ($d$) of an arithmetic progression. We need to find the total number of terms ($n$) and the sum of these terms ($S_n$).

Step 1: Find the number of terms ($n$).
We are given:
First term, $a = 17$
Last term, $a_n = 350$
Common difference, $d = 9$

We can use the formula for the nth term of an AP: $a_n = a + (n-1)d$.
Substitute the given values into the formula:
$350 = 17 + (n-1)9$

Now, we need to solve this equation for $n$.
Subtract 17 from both sides:
$350 – 17 = (n-1)9$
$333 = (n-1)9$

Divide both sides by 9:
$\frac{333}{9} = n-1$
$37 = n-1$

Add 1 to both sides to find $n$:
$37 + 1 = n$
$n = 38$
So, there are 38 terms in the AP.

Step 2: Find the sum of the terms ($S_n$).
We are given:
Number of terms, $n = 38$
First term, $a = 17$
Last term, $l = 350$

We can use the formula for the sum of an AP: $S_n = \frac{n}{2}(a + l)$.
Substitute the values of $n$, $a$, and $l$ into the formula:
$S_{38} = \frac{38}{2}(17 + 350)$

Calculate the sum:
$S_{38} = 19(367)$
$S_{38} = 6973$

Therefore, there are 38 terms in the AP, and their sum is 6973.

Question:

In an AP given a₃ = 15, S₁₀ = 125, find d and a₁₀.

Concept in a Minute:

The question involves arithmetic progressions (AP). We need to recall the formulas for the nth term of an AP and the sum of the first n terms of an AP.
The nth term of an AP is given by: a_n = a_1 + (n-1)d
The sum of the first n terms of an AP is given by: S_n = n/2 * [2a_1 + (n-1)d] or S_n = n/2 * (a_1 + a_n)
where a_1 is the first term and d is the common difference.

Explanation:

We are given two pieces of information about an AP:
1. The third term (a_3) is 15.
2. The sum of the first 10 terms (S_10) is 125.

We need to find the common difference (d) and the tenth term (a_10).

Step 1: Use the given information to form equations.
From a_3 = 15, we can write:
a_1 + (3-1)d = 15
a_1 + 2d = 15 (Equation 1)

From S_10 = 125, we can use the sum formula:
S_10 = 10/2 * [2a_1 + (10-1)d]
125 = 5 * [2a_1 + 9d]
Divide both sides by 5:
25 = 2a_1 + 9d (Equation 2)

Step 2: Solve the system of equations for a_1 and d.
We have two linear equations with two variables:
Equation 1: a_1 + 2d = 15
Equation 2: 2a_1 + 9d = 25

From Equation 1, we can express a_1 in terms of d:
a_1 = 15 – 2d

Substitute this expression for a_1 into Equation 2:
2(15 – 2d) + 9d = 25
30 – 4d + 9d = 25
30 + 5d = 25
5d = 25 – 30
5d = -5
d = -5 / 5
d = -1

Now that we have the value of d, substitute it back into the expression for a_1:
a_1 = 15 – 2d
a_1 = 15 – 2(-1)
a_1 = 15 + 2
a_1 = 17

Step 3: Find the tenth term (a_10).
Now that we know a_1 = 17 and d = -1, we can find a_10 using the formula for the nth term:
a_n = a_1 + (n-1)d
a_10 = a_1 + (10-1)d
a_10 = 17 + (9)(-1)
a_10 = 17 – 9
a_10 = 8

Therefore, the common difference d is -1 and the tenth term a_10 is 8.

Question:

In an AP given l = 28, S = 144, and there are total 9 terms. Find a.

Concept in a Minute:

This question involves the properties of an Arithmetic Progression (AP). We will use the formulas for the nth term and the sum of the first n terms of an AP.
The nth term of an AP is given by: l = a + (n-1)d
The sum of the first n terms of an AP is given by two formulas:
S = n/2 * (a + l)
S = n/2 * (2a + (n-1)d)
where:
l is the last term
S is the sum of the terms
n is the number of terms
a is the first term
d is the common difference

Explanation:

We are given the last term (l) of an AP, the sum of the terms (S), and the total number of terms (n). We need to find the first term (a).
We can directly use the formula for the sum of the first n terms when the first term (a) and the last term (l) are involved:
S = n/2 * (a + l)

We are given:
l = 28
S = 144
n = 9

Substitute these values into the formula:
144 = 9/2 * (a + 28)

Now, we need to solve this equation for ‘a’.
Multiply both sides by 2:
144 * 2 = 9 * (a + 28)
288 = 9 * (a + 28)

Divide both sides by 9:
288 / 9 = a + 28
32 = a + 28

Subtract 28 from both sides to find ‘a’:
a = 32 – 28
a = 4

Therefore, the first term (a) of the AP is 4.

Step-by-step derivation:
1. Identify the given values: last term (l) = 28, sum of terms (S) = 144, number of terms (n) = 9.
2. Recall the formula for the sum of an AP that includes the first term (a) and the last term (l): S = n/2 * (a + l).
3. Substitute the given values into the formula: 144 = 9/2 * (a + 28).
4. Simplify the equation to solve for ‘a’.
5. Multiply both sides by 2: 288 = 9 * (a + 28).
6. Divide both sides by 9: 32 = a + 28.
7. Subtract 28 from both sides: a = 4.
8. State the final answer: The first term (a) is 4.

Question:

Find the sum of first 40 positive integers divisible by 6.

Concept in a Minute:

The question asks for the sum of an arithmetic progression. An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant. The key concepts are:
1. Identifying the first term (a) of the arithmetic progression.
2. Identifying the common difference (d) of the arithmetic progression.
3. Identifying the number of terms (n) in the arithmetic progression.
4. Using the formula for the sum of the first n terms of an arithmetic progression: Sn = n/2 * [2a + (n-1)d] or Sn = n/2 * (a + l), where l is the last term.

Explanation:

We need to find the sum of the first 40 positive integers divisible by 6.
These numbers form an arithmetic progression: 6, 12, 18, 24, …

1. First term (a): The first positive integer divisible by 6 is 6. So, a = 6.

2. Common difference (d): The difference between consecutive terms divisible by 6 is always 6. So, d = 6.

3. Number of terms (n): We are asked to find the sum of the first 40 positive integers divisible by 6. So, n = 40.

Now, we can use the formula for the sum of the first n terms of an arithmetic progression:
Sn = n/2 * [2a + (n-1)d]

Substitute the values of n, a, and d into the formula:
S40 = 40/2 * [2*6 + (40-1)*6]
S40 = 20 * [12 + (39)*6]
S40 = 20 * [12 + 234]
S40 = 20 * [246]
S40 = 4920

Alternatively, we can find the last term (l) first.
The nth term of an AP is given by an = a + (n-1)d.
The 40th term is a40 = 6 + (40-1)*6 = 6 + 39*6 = 6 + 234 = 240.
Then, using the formula Sn = n/2 * (a + l):
S40 = 40/2 * (6 + 240)
S40 = 20 * (246)
S40 = 4920

Therefore, the sum of the first 40 positive integers divisible by 6 is 4920.

Question:

In an AP, given a = 7, a₁₃ = 35, find d and S₁₃.

Concept in a Minute:

Arithmetic Progression (AP): A sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d).
Formula for the n-th term of an AP: a_n = a + (n-1)d, where a is the first term, n is the term number, and d is the common difference.
Formula for the sum of the first n terms of an AP: S_n = n/2 * [2a + (n-1)d] or S_n = n/2 * (a + a_n), where a is the first term, n is the number of terms, and d is the common difference, and a_n is the n-th term.

Explanation:

The question provides us with the first term (a) and the 13th term (a_13) of an arithmetic progression (AP). We need to find the common difference (d) and the sum of the first 13 terms (S_13).

Step 1: Find the common difference (d).
We know the formula for the n-th term of an AP is a_n = a + (n-1)d.
In this case, n = 13, a = 7, and a_13 = 35.
Substitute these values into the formula:
35 = 7 + (13-1)d
35 = 7 + 12d

Step 2: Solve for d.
Subtract 7 from both sides of the equation:
35 – 7 = 12d
28 = 12d
Divide both sides by 12:
d = 28 / 12
Simplify the fraction:
d = 7 / 3

Step 3: Find the sum of the first 13 terms (S_13).
We can use either of the formulas for the sum of the first n terms of an AP. Since we have the first term (a), the last term (a_13), and the number of terms (n=13), the formula S_n = n/2 * (a + a_n) is more convenient.
Substitute the values:
S_13 = 13/2 * (7 + 35)
S_13 = 13/2 * (42)

Step 4: Calculate S_13.
S_13 = 13 * (42 / 2)
S_13 = 13 * 21
S_13 = 273

Therefore, the common difference (d) is 7/3 and the sum of the first 13 terms (S_13) is 273.

Question:

In an AP Given a₁₂ = 37, d = 3, find a and S₁₂.

Concept in a Minute:

Arithmetic Progression (AP): A sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’.
Formula for the nth term of an AP: a_n = a + (n-1)d, where ‘a’ is the first term, ‘n’ is the term number, and ‘d’ is the common difference.
Formula for the sum of the first ‘n’ terms of an AP: S_n = n/2 [2a + (n-1)d] or S_n = n/2 [a + a_n], where ‘a’ is the first term, ‘a_n’ is the nth term, ‘n’ is the number of terms, and ‘d’ is the common difference.

Explanation:

We are given the 12th term of an arithmetic progression (AP), a_12 = 37, and the common difference, d = 3. We need to find the first term (a) and the sum of the first 12 terms (S_12).

Step 1: Find the first term (a).
We will use the formula for the nth term of an AP: a_n = a + (n-1)d.
Substitute the given values: n = 12, a_12 = 37, and d = 3.
37 = a + (12-1) * 3
37 = a + 11 * 3
37 = a + 33
To find ‘a’, subtract 33 from both sides of the equation:
a = 37 – 33
a = 4

Step 2: Find the sum of the first 12 terms (S_12).
We can use either of the formulas for the sum of the first ‘n’ terms. Let’s use the formula S_n = n/2 [a + a_n] as we have both ‘a’ and ‘a_n’.
Substitute the values: n = 12, a = 4, and a_12 = 37.
S_12 = 12/2 [4 + 37]
S_12 = 6 [41]
S_12 = 246

Alternatively, using the formula S_n = n/2 [2a + (n-1)d]:
Substitute the values: n = 12, a = 4, and d = 3.
S_12 = 12/2 [2*4 + (12-1)*3]
S_12 = 6 [8 + 11*3]
S_12 = 6 [8 + 33]
S_12 = 6 [41]
S_12 = 246

Therefore, the first term (a) is 4 and the sum of the first 12 terms (S_12) is 246.

Question:

Write first four terms of the A.P. when the first term a and the common differenced are given as follows:

a = 4, d = -3

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’. The formula to find the n-th term of an AP is a_n = a + (n-1)d, where ‘a’ is the first term. To find the first four terms, we can use this formula or simply add the common difference to the previous term.

Explanation:

The question asks for the first four terms of an Arithmetic Progression (AP) given the first term (a) and the common difference (d).
Given:
First term, a = 4
Common difference, d = -3

To find the first four terms, we can use the definition of an AP, where each subsequent term is obtained by adding the common difference to the preceding term.

The first term is already given:
First term (a_1) = a = 4

The second term is the first term plus the common difference:
Second term (a_2) = a_1 + d = 4 + (-3) = 4 – 3 = 1

The third term is the second term plus the common difference:
Third term (a_3) = a_2 + d = 1 + (-3) = 1 – 3 = -2

The fourth term is the third term plus the common difference:
Fourth term (a_4) = a_3 + d = -2 + (-3) = -2 – 3 = -5

Alternatively, we can use the formula for the n-th term of an AP: a_n = a + (n-1)d

For the first term (n=1):
a_1 = 4 + (1-1)(-3) = 4 + (0)(-3) = 4 + 0 = 4

For the second term (n=2):
a_2 = 4 + (2-1)(-3) = 4 + (1)(-3) = 4 – 3 = 1

For the third term (n=3):
a_3 = 4 + (3-1)(-3) = 4 + (2)(-3) = 4 – 6 = -2

For the fourth term (n=4):
a_4 = 4 + (4-1)(-3) = 4 + (3)(-3) = 4 – 9 = -5

Therefore, the first four terms of the AP are 4, 1, -2, -5.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

-1.2, -3.2, -5.2, -7.2 …

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we need to find the difference between each pair of consecutive terms. If these differences are the same, it is an AP. If it is an AP, the next terms can be found by adding the common difference to the last term.

Explanation:

Given sequence: -1.2, -3.2, -5.2, -7.2 …

Step 1: Check if the sequence is an AP by finding the difference between consecutive terms.
Difference between the second term and the first term:
d1 = -3.2 – (-1.2) = -3.2 + 1.2 = -2.0

Difference between the third term and the second term:
d2 = -5.2 – (-3.2) = -5.2 + 3.2 = -2.0

Difference between the fourth term and the third term:
d3 = -7.2 – (-5.2) = -7.2 + 5.2 = -2.0

Step 2: Determine if it is an AP.
Since d1 = d2 = d3 = -2.0, the difference between consecutive terms is constant. Therefore, the given sequence is an Arithmetic Progression.

Step 3: Identify the common difference (d).
The common difference, d, is -2.0.

Step 4: Write three more terms.
The last given term is -7.2.
The fifth term will be the fourth term + d:
Fifth term = -7.2 + (-2.0) = -7.2 – 2.0 = -9.2

The sixth term will be the fifth term + d:
Sixth term = -9.2 + (-2.0) = -9.2 – 2.0 = -11.2

The seventh term will be the sixth term + d:
Seventh term = -11.2 + (-2.0) = -11.2 – 2.0 = -13.2

The sequence is an AP.
The common difference, d = -2.0.
The next three terms are -9.2, -11.2, and -13.2.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

$−12,−12,−12,−12$ ….

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we calculate the difference between each pair of consecutive terms. If all these differences are the same, then the sequence is an AP. If it is an AP, we can find the next terms by adding the common difference to the last term.

Explanation:

The given sequence is $−12,−12,−12,−12$ ….

To determine if this is an Arithmetic Progression (AP), we need to check if the difference between consecutive terms is constant.
Let’s calculate the difference between the first and second terms:
$a_2 – a_1 = -12 – (-12) = -12 + 12 = 0$

Now, let’s calculate the difference between the second and third terms:
$a_3 – a_2 = -12 – (-12) = -12 + 12 = 0$

And the difference between the third and fourth terms:
$a_4 – a_3 = -12 – (-12) = -12 + 12 = 0$

Since the difference between each pair of consecutive terms is 0, this sequence is an Arithmetic Progression.
The common difference, $d$, is 0.

To find the next three terms, we continue adding the common difference to the last term.
The last given term is $a_4 = -12$.

The fifth term ($a_5$) will be:
$a_5 = a_4 + d = -12 + 0 = -12$

The sixth term ($a_6$) will be:
$a_6 = a_5 + d = -12 + 0 = -12$

The seventh term ($a_7$) will be:
$a_7 = a_6 + d = -12 + 0 = -12$

So, the given sequence is an AP with a common difference $d=0$. The next three terms are $-12, -12, -12$.

Question:

An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term of the A.P.

Concept in a Minute:

This problem involves Arithmetic Progressions (A.P.). The key concepts are the formula for the n-th term of an A.P. which is $a_n = a + (n-1)d$, where ‘a’ is the first term, ‘n’ is the term number, and ‘d’ is the common difference. We will also use the fact that we can form a system of linear equations using the given information to find the unknown values of ‘a’ and ‘d’.

Explanation:

The problem states that an A.P. has 50 terms.
We are given that the 3rd term ($a_3$) is 12.
Using the formula for the n-th term, we can write this as:
$a_3 = a + (3-1)d$
$12 = a + 2d$ (Equation 1)

We are also given that the last term, which is the 50th term ($a_{50}$), is 106.
Using the formula for the n-th term, we can write this as:
$a_{50} = a + (50-1)d$
$106 = a + 49d$ (Equation 2)

Now we have a system of two linear equations with two variables, ‘a’ and ‘d’:
1) $a + 2d = 12$
2) $a + 49d = 106$

To solve for ‘a’ and ‘d’, we can subtract Equation 1 from Equation 2:
$(a + 49d) – (a + 2d) = 106 – 12$
$a + 49d – a – 2d = 94$
$47d = 94$
$d = 94 / 47$
$d = 2$

Now that we have the common difference ‘d’, we can substitute it back into Equation 1 to find the first term ‘a’:
$a + 2d = 12$
$a + 2(2) = 12$
$a + 4 = 12$
$a = 12 – 4$
$a = 8$

So, the first term of the A.P. is 8 and the common difference is 2.

The question asks for the 29th term of the A.P. ($a_{29}$).
Using the formula for the n-th term:
$a_{29} = a + (29-1)d$
$a_{29} = 8 + (28) * 2$
$a_{29} = 8 + 56$
$a_{29} = 64$

Therefore, the 29th term of the A.P. is 64.

Question:

In the following APs, find the missing term in the box:

2, $□$, 26

Concept in a Minute:

Arithmetic Progression (AP): An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’. The general form of an AP is a, a+d, a+2d, a+3d, …
Formula for nth term of an AP: The nth term of an AP is given by a_n = a + (n-1)d, where ‘a’ is the first term and ‘d’ is the common difference.

Explanation:

The given sequence is an arithmetic progression (AP): 2, □, 26.
This means that the difference between consecutive terms is constant. Let the missing term be represented by ‘x’.
So, the sequence is 2, x, 26.
According to the property of an AP, the difference between the second term and the first term is equal to the difference between the third term and the second term.
Therefore, x – 2 = 26 – x.

To solve for x, we can rearrange the equation:
Add ‘x’ to both sides:
x – 2 + x = 26 – x + x
2x – 2 = 26

Add ‘2’ to both sides:
2x – 2 + 2 = 26 + 2
2x = 28

Divide both sides by ‘2’:
2x / 2 = 28 / 2
x = 14

Alternatively, we can use the formula for the nth term.
In this AP, the first term (a_1) is 2.
The third term (a_3) is 26.
Let the missing term be the second term (a_2).
The formula for the nth term is a_n = a + (n-1)d.
For the third term: a_3 = a_1 + (3-1)d
26 = 2 + 2d
Subtract 2 from both sides:
26 – 2 = 2d
24 = 2d
Divide by 2:
d = 12

Now we can find the second term (a_2) using the formula:
a_2 = a_1 + (2-1)d
a_2 = 2 + (1) * 12
a_2 = 2 + 12
a_2 = 14

The missing term is 14.

Question:

Find the sum given below:

–5 + (–8) + (–11) + … + (–230)

Concept in a Minute:

This problem involves finding the sum of an arithmetic progression (AP). An AP is a sequence of numbers such that the difference between the consecutive terms is constant. The key formulas needed are:
1. To find the number of terms (n) in an AP: $a_n = a_1 + (n-1)d$, where $a_n$ is the last term, $a_1$ is the first term, and $d$ is the common difference.
2. To find the sum of an AP (S_n): $S_n = \frac{n}{2}(a_1 + a_n)$ or $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.

Explanation:

The given sequence is –5 + (–8) + (–11) + … + (–230).
This is an arithmetic progression because the difference between consecutive terms is constant.
The first term ($a_1$) is –5.
The common difference ($d$) can be found by subtracting any term from its succeeding term. For example, $d = -8 – (-5) = -8 + 5 = -3$. Also, $-11 – (-8) = -11 + 8 = -3$.
The last term ($a_n$) is –230.

Step 1: Find the number of terms (n) in the AP.
Using the formula $a_n = a_1 + (n-1)d$:
–230 = –5 + (n-1)(–3)
–230 + 5 = (n-1)(–3)
–225 = (n-1)(–3)
Divide both sides by –3:
$\frac{-225}{-3}$ = n-1
75 = n-1
n = 75 + 1
n = 76

So, there are 76 terms in the AP.

Step 2: Find the sum of the AP.
Using the formula $S_n = \frac{n}{2}(a_1 + a_n)$:
$S_{76} = \frac{76}{2}(-5 + (-230))$
$S_{76} = 38(-5 – 230)$
$S_{76} = 38(-235)$

Now, calculate the product:
38 * 235
(40 – 2) * 235 = 40 * 235 – 2 * 235
40 * 235 = 9400
2 * 235 = 470
9400 – 470 = 8930

Since the terms are negative, the sum will be negative.
$S_{76} = -8930$

Therefore, the sum of the given AP is –8930.

Question:

In the following situation, involved make an arithmetic progression? and why?

The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). To check if a situation forms an AP, we need to see if the difference between any two consecutive terms is the same.

Explanation:

Let’s analyze the cost of digging the well for each meter.
The cost for the 1st meter is Rs 150.
The cost for the 2nd meter is Rs 150 (cost of the first meter) + Rs 50 (increase for the subsequent meter) = Rs 200.
The cost for the 3rd meter is Rs 200 (cost of the second meter) + Rs 50 (increase for the subsequent meter) = Rs 250.
The cost for the 4th meter is Rs 250 (cost of the third meter) + Rs 50 (increase for the subsequent meter) = Rs 300.

We can see a sequence of costs: 150, 200, 250, 300, …

Now, let’s check the difference between consecutive terms:
2nd term – 1st term = 200 – 150 = 50
3rd term – 2nd term = 250 – 200 = 50
4th term – 3rd term = 300 – 250 = 50

Since the difference between each consecutive term is constant (50), this situation forms an arithmetic progression.

Why:
The problem states that the cost rises by Rs 50 for each subsequent metre. This means that the increase in cost from one metre to the next is always the same, which is the defining characteristic of an arithmetic progression. The first term (a) of the AP is the cost of digging the first metre (Rs 150), and the common difference (d) is the additional cost for each subsequent metre (Rs 50).

Question:

Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^thterm?

Concept in a Minute:

Arithmetic Progression (A.P.), General term of an A.P., Finding the term number.

Explanation:

Let the given A.P. be denoted by $a_1, a_2, a_3, \ldots$.
The first term ($a$) is 3.
The common difference ($d$) can be found by subtracting any term from its succeeding term.
$d = a_2 – a_1 = 15 – 3 = 12$.
$d = a_3 – a_2 = 27 – 15 = 12$.
So, the common difference is $d = 12$.

The general term of an A.P. is given by the formula:
$a_n = a + (n-1)d$

We need to find which term of the A.P. will be 132 more than its 54th term.
First, let’s find the 54th term ($a_{54}$).
Using the general term formula with $n=54$:
$a_{54} = a + (54-1)d$
$a_{54} = 3 + (53) \times 12$
$a_{54} = 3 + 636$
$a_{54} = 639$

Now, we are looking for a term, let’s call it the $n$th term ($a_n$), which is 132 more than the 54th term.
So, $a_n = a_{54} + 132$.
Substitute the value of $a_{54}$:
$a_n = 639 + 132$
$a_n = 771$

Now we need to find the term number ($n$) for which $a_n = 771$.
Using the general term formula again:
$a_n = a + (n-1)d$
Substitute the values of $a_n$, $a$, and $d$:
$771 = 3 + (n-1)12$

Now, solve for $n$:
Subtract 3 from both sides:
$771 – 3 = (n-1)12$
$768 = (n-1)12$

Divide both sides by 12:
$\frac{768}{12} = n-1$
$64 = n-1$

Add 1 to both sides:
$64 + 1 = n$
$n = 65$

Therefore, the 65th term of the A.P. will be 132 more than its 54th term.

Final Answer Check:
54th term = 639
65th term = $a + (65-1)d = 3 + (64) \times 12 = 3 + 768 = 771$
Difference = 771 – 639 = 132.
The answer is correct.

Question:

Determine the A.P. whose 3^rd term is 16 and the 7^th term exceeds the 5^th term by 12.

Concept in a Minute:

This question involves Arithmetic Progressions (A.P.). The key concepts needed are:
1. The general form of an A.P.: a, a+d, a+2d, …, where ‘a’ is the first term and ‘d’ is the common difference.
2. The formula for the n-th term of an A.P.: a_n = a + (n-1)d.
3. How to set up and solve a system of linear equations to find the unknown values of ‘a’ and ‘d’.

Explanation:

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’.
The n-th term of an A.P. is given by the formula: a_n = a + (n-1)d.

From the given information:
1. The 3rd term is 16.
This can be written as: a_3 = a + (3-1)d = a + 2d.
So, we have the equation: a + 2d = 16 (Equation 1).

2. The 7th term exceeds the 5th term by 12.
This means: a_7 – a_5 = 12.
Using the formula for the n-th term:
a_7 = a + (7-1)d = a + 6d.
a_5 = a + (5-1)d = a + 4d.
Substituting these into the difference equation:
(a + 6d) – (a + 4d) = 12.
a + 6d – a – 4d = 12.
2d = 12.
d = 12 / 2.
d = 6.

Now that we have the common difference (d=6), we can substitute this value into Equation 1 to find the first term ‘a’.
a + 2d = 16.
a + 2(6) = 16.
a + 12 = 16.
a = 16 – 12.
a = 4.

So, the first term is 4 and the common difference is 6.
The A.P. is: 4, 4+6, 4+2(6), …
Which is: 4, 10, 16, …

To verify:
3rd term = 16 (which is correct).
7th term = a + 6d = 4 + 6(6) = 4 + 36 = 40.
5th term = a + 4d = 4 + 4(6) = 4 + 24 = 28.
Difference between 7th and 5th term = 40 – 28 = 12 (which is correct).

The A.P. is 4, 10, 16, 22, 28, 34, 40, …

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

$\sqrt{3,\sqrt{6,\sqrt{9,\sqrt{12…}$

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we calculate the difference between consecutive terms. If these differences are all the same, the sequence is an AP. If it is an AP, we can find the next terms by adding the common difference to the last term.

Explanation:

First, let’s write down the given sequence:
$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, …$

To determine if this is an Arithmetic Progression (AP), we need to check if the difference between consecutive terms is constant.

Calculate the difference between the second term and the first term:
$a_2 – a_1 = \sqrt{6} – \sqrt{3}$

Calculate the difference between the third term and the second term:
$a_3 – a_2 = \sqrt{9} – \sqrt{6} = 3 – \sqrt{6}$

Calculate the difference between the fourth term and the third term:
$a_4 – a_3 = \sqrt{12} – \sqrt{9} = \sqrt{4 \times 3} – 3 = 2\sqrt{3} – 3$

Now, let’s compare these differences:
$\sqrt{6} – \sqrt{3}$
$3 – \sqrt{6}$
$2\sqrt{3} – 3$

It is clear that these differences are not equal.
$\sqrt{6} – \sqrt{3} \neq 3 – \sqrt{6}$
$\sqrt{6} – \sqrt{3} \approx 2.449 – 1.732 = 0.717$
$3 – \sqrt{6} \approx 3 – 2.449 = 0.551$

Since the differences between consecutive terms are not constant, the given sequence is not an Arithmetic Progression.

Answer: The given sequence is not an AP.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

0, -4, -8, -12, …

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we need to calculate the difference between each pair of consecutive terms. If all these differences are the same, then it is an AP. The formula for the n-th term of an AP is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.

Explanation:

The given sequence is 0, -4, -8, -12, …
To determine if this is an Arithmetic Progression (AP), we need to check if the difference between consecutive terms is constant.

Let’s calculate the differences:
Difference between the second term and the first term:
-4 – 0 = -4

Difference between the third term and the second term:
-8 – (-4) = -8 + 4 = -4

Difference between the fourth term and the third term:
-12 – (-8) = -12 + 8 = -4

Since the difference between each consecutive pair of terms is constant and equal to -4, the given sequence is an Arithmetic Progression.

The common difference (d) is -4.

To find the next three terms, we add the common difference to the last known term:
The fifth term will be: -12 + (-4) = -12 – 4 = -16
The sixth term will be: -16 + (-4) = -16 – 4 = -20
The seventh term will be: -20 + (-4) = -20 – 4 = -24

Therefore, the next three terms are -16, -20, and -24.

The final answer is $\boxed{Yes, d = -4, next three terms are -16, -20, -24}$.

Question:

For what value of n, the n^th terms of the arithmetic progressions 63, 65, 67, … and 3, 10, 17, … equal?

Concept in a Minute:

Explanation:

We are given two arithmetic progressions (APs).
AP 1: 63, 65, 67, …
The first term (a₁) is 63.
The common difference (d₁) is 65 – 63 = 2.
The n^th term of this AP is given by a_n = 63 + (n-1)2.

AP 2: 3, 10, 17, …
The first term (a’₁) is 3.
The common difference (d’₁) is 10 – 3 = 7.
The n^th term of this AP is given by a’_n = 3 + (n-1)7.

We need to find the value of n for which the n^th terms of both APs are equal. Therefore, we set the two formulas for the n^th term equal to each other:
63 + (n-1)2 = 3 + (n-1)7

Now, we solve for n:
63 + 2n – 2 = 3 + 7n – 7
61 + 2n = 7n – 4

Subtract 2n from both sides:
61 = 5n – 4

Add 4 to both sides:
65 = 5n

Divide by 5:
n = 65 / 5
n = 13

So, for the 13th term, the two arithmetic progressions will have equal values.

Let’s verify:
For AP 1, the 13th term is 63 + (13-1)2 = 63 + 12 * 2 = 63 + 24 = 87.
For AP 2, the 13th term is 3 + (13-1)7 = 3 + 12 * 7 = 3 + 84 = 87.
The 13th terms are indeed equal.

The final answer is $\boxed{13}$.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

-10, -6, -2, 2 …

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, calculate the difference between each pair of consecutive terms. If these differences are all the same, it’s an AP. If it is an AP, the common difference is this constant value, and the next terms can be found by repeatedly adding the common difference to the last term.

Explanation:

The given sequence is -10, -6, -2, 2 …
To determine if this is an Arithmetic Progression (AP), we need to check if the difference between consecutive terms is constant.

Calculate the difference between the second term and the first term:
d1 = -6 – (-10) = -6 + 10 = 4

Calculate the difference between the third term and the second term:
d2 = -2 – (-6) = -2 + 6 = 4

Calculate the difference between the fourth term and the third term:
d3 = 2 – (-2) = 2 + 2 = 4

Since the differences between consecutive terms are all equal (d1 = d2 = d3 = 4), the given sequence is an Arithmetic Progression.

The common difference, d, is 4.

To find the next three terms, we add the common difference (4) to the last given term (2) repeatedly.

The fifth term is: 2 + 4 = 6
The sixth term is: 6 + 4 = 10
The seventh term is: 10 + 4 = 14

Therefore, the sequence is an AP with a common difference of 4, and the next three terms are 6, 10, and 14.

Question:

In the following APs, find the missing terms in the boxes:

$□,38,□,□,□,−22$

Concept in a Minute:

The key concept is that of an Arithmetic Progression (AP). In an AP, the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’. The nth term of an AP can be found using the formula: a_n = a_1 + (n-1)d, where a_1 is the first term and n is the term number.

Explanation:

The given AP is: $□,38,□,□,□,−22$.
Let the terms of the AP be denoted by $a_1, a_2, a_3, a_4, a_5, a_6$.
We are given:
$a_2 = 38$
$a_6 = -22$

Using the formula for the nth term of an AP, $a_n = a_1 + (n-1)d$:
For $a_2$: $a_2 = a_1 + (2-1)d => 38 = a_1 + d$ (Equation 1)
For $a_6$: $a_6 = a_1 + (6-1)d => -22 = a_1 + 5d$ (Equation 2)

Now we have a system of two linear equations with two variables ($a_1$ and $d$).
Subtract Equation 1 from Equation 2:
$(-22) – 38 = (a_1 + 5d) – (a_1 + d)$
$-60 = 4d$
$d = -60 / 4$
$d = -15$

Now substitute the value of ‘d’ into Equation 1 to find $a_1$:
$38 = a_1 + (-15)$
$38 = a_1 – 15$
$a_1 = 38 + 15$
$a_1 = 53$

Now that we have the first term ($a_1 = 53$) and the common difference ($d = -15$), we can find the missing terms:
$a_1 = 53$
$a_2 = 38$ (given)
$a_3 = a_2 + d = 38 + (-15) = 38 – 15 = 23$
$a_4 = a_3 + d = 23 + (-15) = 23 – 15 = 8$
$a_5 = a_4 + d = 8 + (-15) = 8 – 15 = -7$
$a_6 = -22$ (given)

So, the missing terms are $53, 23, 8, -7$.
The complete AP is: $53, 38, 23, 8, -7, -22$.

The missing terms in the boxes are: $53, 23, 8, -7$.

Question:

In an AP: Given a = 5, d = 3, a_n = 50, find n and S_n.

Concept in a Minute:

Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d).
The formula for the nth term of an AP is a_n = a + (n-1)d, where a is the first term, d is the common difference, and n is the term number.
The formula for the sum of the first n terms of an AP is S_n = n/2 * (a + a_n) or S_n = n/2 * [2a + (n-1)d].

Explanation:

The problem provides us with the first term (a) of an Arithmetic Progression, the common difference (d), and the value of the nth term (a_n). We need to find the term number (n) and the sum of the first n terms (S_n).

Step 1: Identify the given values.
a = 5
d = 3
a_n = 50

Step 2: Use the formula for the nth term of an AP to find n.
a_n = a + (n-1)d
Substitute the given values into the formula:
50 = 5 + (n-1)3

Step 3: Solve the equation for n.
Subtract 5 from both sides:
50 – 5 = (n-1)3
45 = (n-1)3
Divide both sides by 3:
45 / 3 = n-1
15 = n-1
Add 1 to both sides:
15 + 1 = n
n = 16

Step 4: Now that we have found n, use the formula for the sum of the first n terms of an AP to find S_n. We can use either of the two formulas. The first formula is usually easier when a_n is known.
S_n = n/2 * (a + a_n)
Substitute the values of n, a, and a_n into the formula:
S_16 = 16/2 * (5 + 50)

Step 5: Calculate S_n.
S_16 = 8 * (55)
S_16 = 440

Therefore, n = 16 and S_n = 440.

Question:

How many three-digit numbers are divisible by 7?

Concept in a Minute:

The problem requires finding the count of numbers within a specific range (three-digit numbers) that are divisible by a given number (7). This can be solved by identifying the first and last three-digit numbers divisible by 7 and then using arithmetic progression concepts.

Explanation:

A three-digit number ranges from 100 to 999.
We need to find the smallest three-digit number divisible by 7.
Divide 100 by 7: 100 ÷ 7 = 14 with a remainder of 2.
To get the next multiple of 7, add (7 – remainder) to 100. So, 100 + (7 – 2) = 105.
The smallest three-digit number divisible by 7 is 105.

We need to find the largest three-digit number divisible by 7.
Divide 999 by 7: 999 ÷ 7 = 142 with a remainder of 5.
To get the largest multiple of 7 less than or equal to 999, subtract the remainder from 999. So, 999 – 5 = 994.
The largest three-digit number divisible by 7 is 994.

The numbers divisible by 7 form an arithmetic progression with the first term (a) = 105, the last term (l) = 994, and the common difference (d) = 7.
We can use the formula for the nth term of an arithmetic progression: l = a + (n – 1)d, where ‘n’ is the number of terms.
Substituting the values:
994 = 105 + (n – 1)7
Subtract 105 from both sides:
994 – 105 = (n – 1)7
889 = (n – 1)7
Divide both sides by 7:
889 ÷ 7 = n – 1
127 = n – 1
Add 1 to both sides:
n = 127 + 1
n = 128

Therefore, there are 128 three-digit numbers divisible by 7.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

1², 5², 7², 73 …

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we calculate the difference between consecutive terms. If all these differences are the same, then it is an AP. If the sequence is an AP, the next terms can be found by adding the common difference to the last term.

Explanation:

The given sequence is 1^2, 5^2, 7^2, 73 …
First, we need to calculate the actual values of the terms:
1^2 = 1
5^2 = 25
7^2 = 49

So the sequence is 1, 25, 49, 73 …

Now, let’s check if this sequence is an Arithmetic Progression by finding the difference between consecutive terms:
Difference between the second term and the first term: 25 – 1 = 24
Difference between the third term and the second term: 49 – 25 = 24
Difference between the fourth term and the third term: 73 – 49 = 24

Since the difference between consecutive terms is constant (24), the given sequence is an Arithmetic Progression.
The common difference (d) is 24.

To find the next three terms, we add the common difference to the last term:
Fifth term = Fourth term + d = 73 + 24 = 97
Sixth term = Fifth term + d = 97 + 24 = 121
Seventh term = Sixth term + d = 121 + 24 = 145

Therefore, the sequence is an AP with a common difference of 24, and the next three terms are 97, 121, and 145.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

a, 2a, 3a, 4a …

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we calculate the difference between each pair of consecutive terms. If these differences are all the same, it is an AP.

Explanation:

The given sequence is: a, 2a, 3a, 4a, …

To determine if this sequence is an arithmetic progression, we need to check if the difference between consecutive terms is constant.

Let’s calculate the difference between the second term and the first term:
Term 2 – Term 1 = 2a – a = a

Now, let’s calculate the difference between the third term and the second term:
Term 3 – Term 2 = 3a – 2a = a

Next, let’s calculate the difference between the fourth term and the third term:
Term 4 – Term 3 = 4a – 3a = a

Since the difference between each pair of consecutive terms is the same (which is ‘a’), the given sequence is an arithmetic progression.

The common difference (d) is therefore ‘a’.

To find the next three terms, we continue to add the common difference ‘a’ to the last known term.

The last given term is 4a.

The fifth term will be: 4a + d = 4a + a = 5a

The sixth term will be: 5a + d = 5a + a = 6a

The seventh term will be: 6a + d = 6a + a = 7a

So, the sequence is an A.P. with a common difference d = a.
The next three terms are 5a, 6a, and 7a.

Question:

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Concept in a Minute:

This question involves Arithmetic Progressions (AP). We need to use the formulas for the nth term and the sum of the first n terms of an AP.
The formula for the nth term of an AP is $a_n = a + (n-1)d$, where $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference.
The formula for the sum of the first n terms of an AP is $S_n = \frac{n}{2}(a + l)$, where $a$ is the first term, $l$ is the last term, and $n$ is the number of terms.
Alternatively, $S_n = \frac{n}{2}(2a + (n-1)d)$.

Explanation:

We are given:
First term ($a$) = 5
Last term ($l$ or $a_n$) = 45
Sum of the terms ($S_n$) = 400

We need to find:
Number of terms ($n$)
Common difference ($d$)

Step 1: Find the number of terms ($n$) using the sum formula.
We have $S_n = \frac{n}{2}(a + l)$.
Substitute the given values: $400 = \frac{n}{2}(5 + 45)$.
Simplify the equation: $400 = \frac{n}{2}(50)$.
Further simplify: $400 = 25n$.
Solve for $n$: $n = \frac{400}{25}$.
$n = 16$.

Step 2: Find the common difference ($d$) using the nth term formula.
We have $a_n = a + (n-1)d$.
Substitute the known values: $45 = 5 + (16-1)d$.
Simplify the equation: $45 = 5 + 15d$.
Subtract 5 from both sides: $45 – 5 = 15d$.
$40 = 15d$.
Solve for $d$: $d = \frac{40}{15}$.
Simplify the fraction: $d = \frac{8}{3}$.

Therefore, the number of terms is 16 and the common difference is $\frac{8}{3}$.

Question:

If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’. The nth term of an AP can be found using the formula: a_n = a + (n-1)d, where ‘a’ is the first term and ‘n’ is the term number.

Explanation:

Let the first term of the AP be ‘a’ and the common difference be ‘d’.
We are given that the 3rd term (a_3) is 4. Using the formula for the nth term:
a_3 = a + (3-1)d = a + 2d
So, a + 2d = 4 (Equation 1)

We are also given that the 9th term (a_9) is –8. Using the formula for the nth term:
a_9 = a + (9-1)d = a + 8d
So, a + 8d = –8 (Equation 2)

Now we have a system of two linear equations with two variables, ‘a’ and ‘d’. We can solve this system to find the values of ‘a’ and ‘d’.
Subtract Equation 1 from Equation 2:
(a + 8d) – (a + 2d) = –8 – 4
a + 8d – a – 2d = –12
6d = –12
d = –12 / 6
d = –2

Now substitute the value of ‘d’ into Equation 1 to find ‘a’:
a + 2(–2) = 4
a – 4 = 4
a = 4 + 4
a = 8

So, the first term is 8 and the common difference is –2.

We need to find which term of this AP is zero. Let the mth term be zero (a_m = 0).
Using the formula for the nth term:
a_m = a + (m-1)d
0 = 8 + (m-1)(–2)
0 = 8 – 2m + 2
0 = 10 – 2m
2m = 10
m = 10 / 2
m = 5

Therefore, the 5th term of this AP is zero.

Question:

Two APs have the same common difference. The difference between their 100^th term is 100, what is the difference between their 1000^thterms?

Concept in a Minute:

Arithmetic Progression (AP), properties of AP, common difference

Explanation:

Let the two Arithmetic Progressions (APs) be AP1 and AP2.
Let the first term of AP1 be $a_1$ and its common difference be $d$.
Let the first term of AP2 be $a_2$ and its common difference be $d$.
We are given that the common differences of the two APs are the same.

The $n^{th}$ term of an AP is given by the formula: $a_n = a + (n-1)d$

For AP1, the 100th term is:
$a_{100, AP1} = a_1 + (100-1)d = a_1 + 99d$

For AP2, the 100th term is:
$a_{100, AP2} = a_2 + (100-1)d = a_2 + 99d$

We are given that the difference between their 100th terms is 100:
$a_{100, AP1} – a_{100, AP2} = 100$
$(a_1 + 99d) – (a_2 + 99d) = 100$
$a_1 + 99d – a_2 – 99d = 100$
$a_1 – a_2 = 100$

This means the difference between the first terms of the two APs is 100.

Now, we need to find the difference between their 1000th terms.

For AP1, the 1000th term is:
$a_{1000, AP1} = a_1 + (1000-1)d = a_1 + 999d$

For AP2, the 1000th term is:
$a_{1000, AP2} = a_2 + (1000-1)d = a_2 + 999d$

The difference between their 1000th terms is:
$a_{1000, AP1} – a_{1000, AP2} = (a_1 + 999d) – (a_2 + 999d)$
$= a_1 + 999d – a_2 – 999d$
$= a_1 – a_2$

Since we found earlier that $a_1 – a_2 = 100$, the difference between their 1000th terms is also 100.

The final answer is $\boxed{100}$.

Question:

How many multiples of 4 lie between 10 and 250?

Concept in a Minute:

Arithmetic Progression. Multiples of a number form an arithmetic progression. To find the number of terms in an AP, we use the formula: number of terms = (last term – first term)/common difference + 1.

Explanation:

We need to find the number of multiples of 4 that are strictly between 10 and 250. This means the multiples must be greater than 10 and less than 250.

First, find the smallest multiple of 4 that is greater than 10.
We can divide 10 by 4: 10 / 4 = 2 with a remainder of 2.
The next multiple of 4 after 8 (which is 4 * 2) will be 4 * 3 = 12.
So, the first multiple of 4 between 10 and 250 is 12.

Next, find the largest multiple of 4 that is less than 250.
We can divide 250 by 4: 250 / 4 = 62 with a remainder of 2.
This means 250 is not a multiple of 4. The largest multiple of 4 less than 250 would be 250 minus the remainder.
So, the largest multiple of 4 less than 250 is 250 – 2 = 248.

Now we have an arithmetic progression where:
The first term (a) = 12
The last term (l) = 248
The common difference (d) = 4 (since we are looking for multiples of 4)

We can use the formula for the number of terms (n) in an arithmetic progression:
n = (l – a) / d + 1

Substitute the values:
n = (248 – 12) / 4 + 1
n = 236 / 4 + 1
n = 59 + 1
n = 60

Therefore, there are 60 multiples of 4 between 10 and 250.

Question:

In the following situation, involved make an arithmetic progression? and why?

The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference. To determine if a situation forms an AP, we need to check if the difference between successive terms is always the same.

Explanation:

Let’s analyze the taxi fare for each kilometer.
The fare for the first kilometer is ₹ 15.
The fare for the second kilometer is the fare for the first kilometer plus the fare for the additional kilometer: ₹ 15 + ₹ 8 = ₹ 23.
The fare for the third kilometer is the fare for the second kilometer plus the fare for the additional kilometer: ₹ 23 + ₹ 8 = ₹ 31.
The fare for the fourth kilometer is the fare for the third kilometer plus the fare for the additional kilometer: ₹ 31 + ₹ 8 = ₹ 39.

The sequence of taxi fares for each kilometer is 15, 23, 31, 39, …

Now, let’s find the difference between consecutive terms:
Difference between the 2nd and 1st term: 23 – 15 = 8
Difference between the 3rd and 2nd term: 31 – 23 = 8
Difference between the 4th and 3rd term: 39 – 31 = 8

Since the difference between each consecutive term is constant (₹ 8), the sequence of taxi fares forms an arithmetic progression.

The first term (a) of this AP is ₹ 15, and the common difference (d) is ₹ 8.

Question:

Find the sum of first 15 multiples of 8.

Concept in a Minute:

This question involves finding the sum of an arithmetic progression (AP). An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. The sum of the first ‘n’ terms of an AP can be found using the formula: S_n = n/2 * [2a + (n-1)d] or S_n = n/2 * (a + l), where ‘a’ is the first term, ‘d’ is the common difference, ‘n’ is the number of terms, and ‘l’ is the last term.

Explanation:

We need to find the sum of the first 15 multiples of 8.
The multiples of 8 are 8, 16, 24, 32, …
This forms an arithmetic progression where:
The first term (a) = 8
The common difference (d) = 16 – 8 = 8
The number of terms (n) = 15

We can use the formula for the sum of the first ‘n’ terms of an AP:
S_n = n/2 * [2a + (n-1)d]

Substitute the values:
S_15 = 15/2 * [2*8 + (15-1)*8]
S_15 = 15/2 * [16 + (14)*8]
S_15 = 15/2 * [16 + 112]
S_15 = 15/2 * [128]
S_15 = 15 * (128/2)
S_15 = 15 * 64

Now, calculate the product:
15 * 64 = 960

Alternatively, we can first find the 15th multiple of 8 (the last term, l):
l = a + (n-1)d
l = 8 + (15-1)*8
l = 8 + 14*8
l = 8 + 112
l = 120

Then use the formula: S_n = n/2 * (a + l)
S_15 = 15/2 * (8 + 120)
S_15 = 15/2 * (128)
S_15 = 15 * 64
S_15 = 960

Therefore, the sum of the first 15 multiples of 8 is 960.

Question:

In an AP given d = 5, S₉ = 75, find a and a₉.

Concept in a Minute:

Arithmetic Progression (AP) formulas:
Sum of n terms of an AP: S_n = n/2 * [2a + (n-1)d]
nth term of an AP: a_n = a + (n-1)d
where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the number of terms.

Explanation:

We are given the common difference (d = 5) and the sum of the first 9 terms (S_9 = 75) of an arithmetic progression. We need to find the first term (a) and the 9th term (a_9).

Step 1: Use the formula for the sum of n terms of an AP to find the first term ‘a’.
We know S_n = n/2 * [2a + (n-1)d].
Substitute the given values: n = 9, d = 5, S_9 = 75.
75 = 9/2 * [2a + (9-1) * 5]
75 = 9/2 * [2a + 8 * 5]
75 = 9/2 * [2a + 40]

Multiply both sides by 2:
150 = 9 * [2a + 40]

Divide both sides by 9:
150/9 = 2a + 40
50/3 = 2a + 40

Subtract 40 from both sides:
50/3 – 40 = 2a
50/3 – 120/3 = 2a
-70/3 = 2a

Divide both sides by 2:
a = -70 / (3 * 2)
a = -70 / 6
a = -35 / 3

Step 2: Use the formula for the nth term of an AP to find the 9th term ‘a_9’.
We know a_n = a + (n-1)d.
Substitute the values: a = -35/3, n = 9, d = 5.
a_9 = -35/3 + (9-1) * 5
a_9 = -35/3 + 8 * 5
a_9 = -35/3 + 40

To add these, find a common denominator:
a_9 = -35/3 + 120/3
a_9 = (120 – 35) / 3
a_9 = 85 / 3

Therefore, the first term a = -35/3 and the 9th term a_9 = 85/3.

Question:

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Concept in a Minute:

Arithmetic Progression (AP): A sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d).
General term of an AP: The nth term of an AP is given by the formula: a_n = a + (n-1)d, where ‘a’ is the first term and ‘d’ is the common difference.

Explanation:

Let the first term of the Arithmetic Progression be ‘a’ and the common difference be ‘d’.
The 17th term of the AP can be represented as a_17.
Using the general formula for the nth term, a_n = a + (n-1)d:
a_17 = a + (17-1)d
a_17 = a + 16d

The 10th term of the AP can be represented as a_10.
Using the general formula for the nth term, a_n = a + (n-1)d:
a_10 = a + (10-1)d
a_10 = a + 9d

The problem states that the 17th term exceeds its 10th term by 7. This can be written as an equation:
a_17 = a_10 + 7

Now, substitute the expressions for a_17 and a_10 into this equation:
(a + 16d) = (a + 9d) + 7

To find the common difference (d), we need to solve this equation:
a + 16d = a + 9d + 7

Subtract ‘a’ from both sides of the equation:
16d = 9d + 7

Subtract 9d from both sides of the equation:
16d – 9d = 7
7d = 7

Divide both sides by 7:
d = 7 / 7
d = 1

Therefore, the common difference of the AP is 1.

Question:

In the following APs, find the missing term in the boxes:

$−4,□,□,□,□,6$

Concept in a Minute:

This question is about Arithmetic Progressions (APs). An AP is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’. The nth term of an AP can be found using the formula: an = a + (n-1)d, where ‘a’ is the first term and ‘n’ is the term number.

Explanation:

The given sequence is an AP: −4,□,□,□,□,6.
We are given the first term (a) which is -4.
We are also given the sixth term (a6) which is 6.
We can use the formula for the nth term of an AP: an = a + (n-1)d.
For the 6th term, n = 6. So, a6 = a + (6-1)d.
Substituting the given values: 6 = -4 + 5d.
Now, we need to solve for the common difference ‘d’.
Add 4 to both sides of the equation: 6 + 4 = 5d.
This gives us 10 = 5d.
Divide both sides by 5: d = 10 / 5 = 2.
So, the common difference is 2.

Now we can find the missing terms by adding the common difference to the previous term.
The first term is -4.
The second term (□) = -4 + 2 = -2.
The third term (□) = -2 + 2 = 0.
The fourth term (□) = 0 + 2 = 2.
The fifth term (□) = 2 + 2 = 4.
The sixth term is 6, which matches the given information.

Therefore, the missing terms are -2, 0, 2, and 4.
The AP is: -4, -2, 0, 2, 4, 6.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

0.2, 0.22, 0.222, 0.2222 ….

Concept in a Minute:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, calculate the difference between each pair of consecutive terms. If all these differences are the same, then the sequence is an AP.

Explanation:

To determine if the sequence 0.2, 0.22, 0.222, 0.2222 … is an Arithmetic Progression (AP), we need to check if the difference between consecutive terms is constant.

Let’s calculate the differences:
First term (a1) = 0.2
Second term (a2) = 0.22
Third term (a3) = 0.222
Fourth term (a4) = 0.2222

Difference between the second and first term:
d1 = a2 – a1 = 0.22 – 0.2 = 0.02

Difference between the third and second term:
d2 = a3 – a2 = 0.222 – 0.22 = 0.002

Difference between the fourth and third term:
d3 = a4 – a3 = 0.2222 – 0.222 = 0.0002

We observe that the differences between consecutive terms (d1, d2, d3) are not the same: 0.02 ≠ 0.002 ≠ 0.0002.
Since the difference between consecutive terms is not constant, the given sequence is not an Arithmetic Progression.

Therefore, we cannot find a common difference (d) or write three more terms as it does not form an AP.

Question:

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Concept in a Minute:

This problem involves arithmetic progression (AP). An AP is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference. The formula for the nth term of an AP is $a_n = a + (n-1)d$, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number.

Explanation:

We can model Subba Rao’s annual salary as an arithmetic progression.
The first term (a) is his starting salary in 1995, which is Rs 5000.
The common difference (d) is the annual increment, which is Rs 200.
We want to find the year when his income reached Rs 7000. Let this be the nth term ($a_n$).
So, $a_n = 7000$.

Using the formula for the nth term of an AP:
$a_n = a + (n-1)d$
Substitute the given values:
$7000 = 5000 + (n-1)200$

Now, we solve for ‘n’:
$7000 – 5000 = (n-1)200$
$2000 = (n-1)200$
Divide both sides by 200:
$2000 / 200 = n-1$
$10 = n-1$
$n = 10 + 1$
$n = 11$

This means that his income reached Rs 7000 in the 11th year of his service.
Since he started in 1995, the 11th year can be calculated as:
Year = Starting Year + (n – 1)
Year = 1995 + (11 – 1)
Year = 1995 + 10
Year = 2005

Therefore, Subba Rao’s income reached Rs 7000 in the year 2005.

Question:

Write first four terms of the A.P. when the first term a and the common difference d are given as follows:

a = -1.25, d = -0.25

Concept in a Minute:

An arithmetic progression (A.P.) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). The general form of an A.P. is a, a+d, a+2d, a+3d, … where ‘a’ is the first term.

Explanation:

The question asks for the first four terms of an Arithmetic Progression (A.P.). We are given the first term (a) and the common difference (d).

Given:
First term, a = -1.25
Common difference, d = -0.25

To find the first four terms, we use the formula for the nth term of an A.P., which is $a_n = a + (n-1)d$. However, for the first few terms, it’s simpler to just add the common difference iteratively.

The first term is given:
$a_1 = a = -1.25$

The second term is found by adding the common difference to the first term:
$a_2 = a_1 + d = -1.25 + (-0.25) = -1.25 – 0.25 = -1.50$

The third term is found by adding the common difference to the second term:
$a_3 = a_2 + d = -1.50 + (-0.25) = -1.50 – 0.25 = -1.75$

The fourth term is found by adding the common difference to the third term:
$a_4 = a_3 + d = -1.75 + (-0.25) = -1.75 – 0.25 = -2.00$

Therefore, the first four terms of the A.P. are -1.25, -1.50, -1.75, and -2.00.

Question:

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Concept in a Minute:

This question involves Arithmetic Progression (AP). The key concepts are:
1. The formula for the nth term of an AP: a_n = a + (n-1)d, where ‘a’ is the first term and ‘d’ is the common difference.
2. The formula for the sum of the first ‘n’ terms of an AP: S_n = (n/2) * [2a + (n-1)d] or S_n = (n/2) * (a + l), where ‘l’ is the last term.

Explanation:

We are given the second term (a_2) and the third term (a_3) of an AP.
a_2 = 14
a_3 = 18

The common difference ‘d’ of an AP is the difference between any two consecutive terms.
So, d = a_3 – a_2
d = 18 – 14
d = 4

Now we know the common difference. We can find the first term ‘a’ using the formula for the nth term.
a_2 = a + (2-1)d
14 = a + 1*4
14 = a + 4
a = 14 – 4
a = 10

So, the first term is 10 and the common difference is 4.

We need to find the sum of the first 51 terms (S_51). We will use the formula:
S_n = (n/2) * [2a + (n-1)d]

Here, n = 51, a = 10, and d = 4.
S_51 = (51/2) * [2*10 + (51-1)*4]
S_51 = (51/2) * [20 + (50)*4]
S_51 = (51/2) * [20 + 200]
S_51 = (51/2) * [220]
S_51 = 51 * (220/2)
S_51 = 51 * 110

Now, calculate 51 * 110:
51 * 110 = 51 * (100 + 10)
= 51 * 100 + 51 * 10
= 5100 + 510
= 5610

Therefore, the sum of the first 51 terms is 5610.

Question:

Find the sum of the following APs.

$115,112,110$, ……, to 11 terms.

Concept in a Minute:

Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. This constant difference is called the common difference (d).
Sum of an AP: The sum of the first ‘n’ terms of an AP can be calculated using the formula: $S_n = n/2 * [2a + (n-1)d]$ or $S_n = n/2 * (a + l)$, where ‘a’ is the first term, ‘d’ is the common difference, ‘n’ is the number of terms, and ‘l’ is the last term.

Explanation:

The given sequence is an arithmetic progression (AP): $115, 112, 110$, ……, to 11 terms.

Step 1: Identify the first term (a).
The first term of the AP is $a = 115$.

Step 2: Identify the common difference (d).
To find the common difference, subtract any term from its succeeding term.
$d = 112 – 115 = -3$
$d = 110 – 112 = -2$
Wait! The difference is not constant. Let’s re-examine the sequence.

The sequence given is $115, 112, 110$.
The difference between the first and second term is $112 – 115 = -3$.
The difference between the second and third term is $110 – 112 = -2$.

Since the difference between consecutive terms is not constant, this sequence is not an arithmetic progression as stated in the question.

Therefore, the question cannot be answered as it is presented.

Question:

Find the number of terms in the following A.P.: 7, 13, 19, …, 205.

Concept in a Minute:

The problem involves an Arithmetic Progression (A.P.). An A.P. is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). The formula to find the nth term (a_n) of an A.P. is a_n = a_1 + (n-1)d, where a_1 is the first term and n is the number of terms. To find the number of terms, we will use this formula and solve for n.

Explanation:

The given sequence is 7, 13, 19, …, 205.
This is an Arithmetic Progression (A.P.).

First term (a_1) = 7.
To find the common difference (d), subtract any term from its succeeding term.
d = 13 – 7 = 6.
d = 19 – 13 = 6.
So, the common difference (d) is 6.

The last term of the A.P. is given as 205. Let this be the nth term (a_n).
So, a_n = 205.

We use the formula for the nth term of an A.P.:
a_n = a_1 + (n-1)d

Substitute the known values into the formula:
205 = 7 + (n-1)6

Now, we need to solve for n (the number of terms).
Subtract 7 from both sides of the equation:
205 – 7 = (n-1)6
198 = (n-1)6

Divide both sides by 6:
198 / 6 = n-1
33 = n-1

Add 1 to both sides to find n:
33 + 1 = n
n = 34

Therefore, there are 34 terms in the given A.P.

Question:

In an AP given a = 8, a_n = 62, S_n = 210, find n and d.

Concept in a Minute:

The question involves arithmetic progressions (AP). The key concepts are the formulas for the nth term and the sum of the first n terms of an AP.
1. nth term of an AP: $a_n = a + (n-1)d$
2. Sum of the first n terms of an AP: $S_n = n/2 * (a + a_n)$ or $S_n = n/2 * (2a + (n-1)d)$

Explanation:

We are given the first term ($a$), the nth term ($a_n$), and the sum of the first n terms ($S_n$) of an arithmetic progression. We need to find the number of terms ($n$) and the common difference ($d$).

Step 1: Use the formula for the sum of an AP to find ‘n’.
We know $S_n = n/2 * (a + a_n)$.
Substitute the given values: $210 = n/2 * (8 + 62)$.
Simplify the equation: $210 = n/2 * (70)$.
Further simplify: $210 = 35n$.
Solve for n: $n = 210 / 35$.
Calculate n: $n = 6$.

Step 2: Use the formula for the nth term of an AP to find ‘d’.
We know $a_n = a + (n-1)d$.
Substitute the values of $a_n$, $a$, and the calculated value of $n$: $62 = 8 + (6-1)d$.
Simplify the equation: $62 = 8 + 5d$.
Subtract 8 from both sides: $62 – 8 = 5d$.
This gives: $54 = 5d$.
Solve for d: $d = 54 / 5$.
Calculate d: $d = 10.8$.

Therefore, the number of terms $n = 6$ and the common difference $d = 10.8$.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

1², 3², 5², 7² …

Concept in a Minute:

Arithmetic Progression (AP): A sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d).
To check if a sequence is an AP, we calculate the difference between each pair of consecutive terms. If all these differences are the same, then it is an AP.

Explanation:

The given sequence is 1², 3², 5², 7² …

First, let’s calculate the actual values of these terms:
Term 1: 1² = 1
Term 2: 3² = 9
Term 3: 5² = 25
Term 4: 7² = 49

So the sequence is 1, 9, 25, 49, …

Now, we need to check if this sequence is an Arithmetic Progression by finding the difference between consecutive terms:

Difference between Term 2 and Term 1:
d₁ = 9 – 1 = 8

Difference between Term 3 and Term 2:
d₂ = 25 – 9 = 16

Difference between Term 4 and Term 3:
d₃ = 49 – 25 = 24

We can see that the differences between consecutive terms are not constant (8 ≠ 16 ≠ 24).

Therefore, the given sequence 1², 3², 5², 7² … is not an AP.

Since the question asks to find the common difference ‘d’ and write three more terms *if* they form an A.P., and we have determined that it does not form an A.P., we stop here.

Question:

Find the sum of the following APs:

2, 7, 12, …, to 10 terms.

Concept in a Minute:

The question asks to find the sum of an Arithmetic Progression (AP). An AP is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference. The formula to find the sum of the first ‘n’ terms of an AP is given by:
Sn = n/2 [2a + (n-1)d]
where:
Sn = sum of the first n terms
n = number of terms
a = first term
d = common difference

Explanation:

The given Arithmetic Progression (AP) is 2, 7, 12, … and we need to find the sum of its first 10 terms.
First, identify the first term (a) and the common difference (d) of the AP.
The first term, a = 2.
The common difference, d, is the difference between any two consecutive terms.
d = 7 – 2 = 5
d = 12 – 7 = 5
So, the common difference d = 5.
The number of terms, n = 10.

Now, use the formula for the sum of an AP:
Sn = n/2 [2a + (n-1)d]

Substitute the values of n, a, and d into the formula:
S10 = 10/2 [2(2) + (10-1)5]
S10 = 5 [4 + (9)5]
S10 = 5 [4 + 45]
S10 = 5 [49]
S10 = 245

Therefore, the sum of the first 10 terms of the AP 2, 7, 12, … is 245.

Question:

Find the sum of the following APs.

0.6, 1.7, 2.8, …….., to 100 terms.

Concept in a Minute:

The question asks to find the sum of an arithmetic progression (AP). An AP is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). The sum of the first ‘n’ terms of an AP can be calculated using the formula: Sn = n/2 * [2a + (n-1)d], where ‘a’ is the first term and ‘n’ is the number of terms.

Explanation:

The given sequence is an arithmetic progression: 0.6, 1.7, 2.8, …
The first term (a) is 0.6.
The common difference (d) is the difference between any two consecutive terms.
d = 1.7 – 0.6 = 1.1
d = 2.8 – 1.7 = 1.1
So, the common difference (d) is 1.1.
The number of terms (n) is given as 100.

We need to find the sum of the first 100 terms (S100).
Using the formula for the sum of an AP: Sn = n/2 * [2a + (n-1)d]

Substitute the values of a, d, and n into the formula:
S100 = 100/2 * [2 * 0.6 + (100-1) * 1.1]
S100 = 50 * [1.2 + (99) * 1.1]
S100 = 50 * [1.2 + 108.9]
S100 = 50 * [110.1]
S100 = 5505

Therefore, the sum of the first 100 terms of the AP is 5505.

Question:

Write first four terms of the A.P. when the first term a and the common differenced are given as follows:

a = 10, d = 10

Concept in a Minute:

An arithmetic progression (A.P.) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). The formula to find the nth term of an A.P. is a_n = a + (n-1)d, where ‘a’ is the first term and ‘d’ is the common difference. To find the first four terms, we use the first term and repeatedly add the common difference.

Explanation:

We are given the first term (a) of an arithmetic progression as 10 and the common difference (d) as 10.
The first term is already given, which is a_1 = a = 10.
To find the second term (a_2), we add the common difference to the first term: a_2 = a_1 + d = 10 + 10 = 20.
To find the third term (a_3), we add the common difference to the second term: a_3 = a_2 + d = 20 + 10 = 30.
To find the fourth term (a_4), we add the common difference to the third term: a_4 = a_3 + d = 30 + 10 = 40.
Therefore, the first four terms of the A.P. are 10, 20, 30, and 40.

The first four terms of the A.P. are 10, 20, 30, 40.

Question:

In an AP given a = 3, n = 8, S_n = 192, find d.

Concept in a Minute:

The question involves an arithmetic progression (AP). The key concepts needed are the formula for the sum of the first ‘n’ terms of an AP and the relationship between the first term, common difference, number of terms, and the sum.
The formula for the sum of the first ‘n’ terms of an AP is given by:
S_n = n/2 * [2a + (n-1)d]
where:
S_n is the sum of the first ‘n’ terms
n is the number of terms
a is the first term
d is the common difference

Explanation:

We are given:
The first term, a = 3
The number of terms, n = 8
The sum of the first n terms, S_n = 192
We need to find the common difference, d.

We can use the formula for the sum of an AP:
S_n = n/2 * [2a + (n-1)d]

Substitute the given values into the formula:
192 = 8/2 * [2(3) + (8-1)d]

Simplify the equation:
192 = 4 * [6 + 7d]

Now, distribute the 4:
192 = 24 + 28d

To isolate the term with ‘d’, subtract 24 from both sides:
192 – 24 = 28d
168 = 28d

Finally, solve for ‘d’ by dividing both sides by 28:
d = 168 / 28
d = 6

Therefore, the common difference d is 6.

Question:

Which term of the A.P. 3, 8, 13, 18, …, is 78?

Concept in a Minute:

This question involves arithmetic progressions (A.P.). An A.P. is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference. The formula to find the nth term of an A.P. is given by: a_n = a + (n-1)d, where a_n is the nth term, a is the first term, n is the term number, and d is the common difference.

Explanation:

The given A.P. is 3, 8, 13, 18, …
First, identify the first term (a) and the common difference (d).
The first term, a = 3.
The common difference, d = 8 – 3 = 5 (or 13 – 8 = 5, etc.).
We are asked to find which term is 78. Let this be the nth term, so a_n = 78.
Now, we use the formula for the nth term of an A.P.: a_n = a + (n-1)d.
Substitute the known values into the formula: 78 = 3 + (n-1)5.
Now, we need to solve for n.
Subtract 3 from both sides: 78 – 3 = (n-1)5.
75 = (n-1)5.
Divide both sides by 5: 75 / 5 = n-1.
15 = n-1.
Add 1 to both sides: 15 + 1 = n.
n = 16.
Therefore, the 16th term of the A.P. is 78.

Question:

Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n^thweek, her week, her weekly savings become Rs 20.75, find n.

Concept in a Minute:

This question involves arithmetic progression (AP). An AP is a sequence of numbers such that the difference between the consecutive terms is constant. The formula for the n-th term of an AP is given by: a_n = a + (n-1)d, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number.

Explanation:

Ramkali’s weekly savings form an arithmetic progression.
The first week’s saving (first term, a) = Rs 5.
The increase in weekly saving (common difference, d) = Rs 1.75.
Her weekly savings in the n-th week (n-th term, a_n) = Rs 20.75.
We need to find the number of weeks (n).

Using the formula for the n-th term of an AP:
a_n = a + (n-1)d

Substitute the given values into the formula:
20.75 = 5 + (n-1)1.75

Now, we need to solve for ‘n’:
Subtract 5 from both sides:
20.75 – 5 = (n-1)1.75
15.75 = (n-1)1.75

Divide both sides by 1.75:
15.75 / 1.75 = n-1
9 = n-1

Add 1 to both sides:
9 + 1 = n
10 = n

Therefore, in the 10th week, her weekly savings become Rs 20.75.

The value of n is 10.

Question:

Find the sum of the odd numbers between 0 and 50.

Concept in a Minute:

This problem requires understanding of arithmetic progressions (AP). Specifically, you need to identify the first term (a), the last term (l), and the common difference (d) of the AP formed by odd numbers. Then, you’ll need to find the number of terms (n) in this AP and finally use the formula for the sum of an AP, which is S_n = n/2 * (a + l).

Explanation:

The odd numbers between 0 and 50 are 1, 3, 5, …, 49.
This forms an arithmetic progression with:
First term (a) = 1
Common difference (d) = 2 (since consecutive odd numbers differ by 2)
Last term (l) = 49

To find the number of terms (n), we can use the formula for the nth term of an AP: l = a + (n-1)d.
Substituting the values: 49 = 1 + (n-1)2
49 – 1 = (n-1)2
48 = 2(n-1)
48 / 2 = n-1
24 = n-1
n = 24 + 1
n = 25

Now, we find the sum of this AP (S_n) using the formula: S_n = n/2 * (a + l).
S_25 = 25/2 * (1 + 49)
S_25 = 25/2 * 50
S_25 = 25 * (50 / 2)
S_25 = 25 * 25
S_25 = 625

Therefore, the sum of the odd numbers between 0 and 50 is 625.

Question:

Find the sum of the following APs.

−37, −33, −29, …, to 12 terms.

Concept in a Minute:

Arithmetic Progression (AP) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference (d). The formula to find the sum of the first n terms of an AP is given by S_n = n/2 * [2a + (n-1)d], where S_n is the sum of n terms, n is the number of terms, a is the first term, and d is the common difference.

Explanation:

The given sequence is an Arithmetic Progression (AP) because the difference between consecutive terms is constant.
The first term (a) is -37.
The common difference (d) can be found by subtracting any term from its succeeding term. For example, d = -33 – (-37) = -33 + 37 = 4. Similarly, d = -29 – (-33) = -29 + 33 = 4.
We need to find the sum of the first 12 terms, so n = 12.
Using the formula for the sum of an AP, S_n = n/2 * [2a + (n-1)d]:
S_12 = 12/2 * [2*(-37) + (12-1)*4]
S_12 = 6 * [-74 + (11)*4]
S_12 = 6 * [-74 + 44]
S_12 = 6 * [-30]
S_12 = -180.

The sum of the first 12 terms of the given AP is -180.

Question:

Following are APs or not? If they form an A.P. find the common difference d and write three more terms:

2, 4, 8, 16 …

Concept in a Minute:

An arithmetic progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). To check if a sequence is an AP, we calculate the difference between each pair of consecutive terms. If all these differences are the same, then it is an AP.

Explanation:

To determine if the sequence 2, 4, 8, 16 … is an arithmetic progression (AP), we need to check the difference between consecutive terms.

Difference between the second term and the first term:
4 – 2 = 2

Difference between the third term and the second term:
8 – 4 = 4

Difference between the fourth term and the third term:
16 – 8 = 8

Since the differences between consecutive terms (2, 4, and 8) are not constant, the given sequence 2, 4, 8, 16 … is not an arithmetic progression.

Therefore, we cannot find a common difference ‘d’ and write three more terms as it does not form an AP.

Question:

Check whether -150 is a term of the A.P. 11, 8, 5, 2, ….

Concept in a Minute:

The general form of an arithmetic progression (A.P.) is given by a_n = a + (n-1)d, where a_n is the n-th term, a is the first term, n is the term number, and d is the common difference. To check if a number is a term of an A.P., we assume it is the n-th term and solve for n. If n is a positive integer, then the number is a term of the A.P.

Explanation:

The given A.P. is 11, 8, 5, 2, ….
The first term, a = 11.
The common difference, d = 8 – 11 = -3.
We need to check if -150 is a term of this A.P. Let’s assume -150 is the n-th term (a_n).
Using the formula for the n-th term of an A.P., a_n = a + (n-1)d:
-150 = 11 + (n-1)(-3)
-150 – 11 = (n-1)(-3)
-161 = -3(n-1)
Divide both sides by -3:
-161 / -3 = n-1
161 / 3 = n-1
Now, calculate 161 / 3:
161 divided by 3 is not a whole number. 161 / 3 = 53.666…
So, n-1 = 53.666…
n = 53.666… + 1
n = 54.666…
Since n is not a positive integer, -150 is not a term of the given A.P.

Question:

For the following APs, write the first term and the common difference:

3, 1, –1, –3, …

Concept in a Minute:

Explanation:

The given arithmetic progression (AP) is: 3, 1, –1, –3, …

To find the first term, we identify the very first number in the given sequence.
First term (a) = 3

To find the common difference (d), we subtract any term from its succeeding term. Let’s take the first two terms:
d = Second term – First term
d = 1 – 3
d = –2

Let’s verify with the next pair of terms:
d = Third term – Second term
d = –1 – 1
d = –2

And with the next pair:
d = Fourth term – Third term
d = –3 – (–1)
d = –3 + 1
d = –2

Since the difference between consecutive terms is constant, the common difference is indeed –2.

Therefore,
The first term is 3.
The common difference is –2.

Question:

For the following A.P.s, write the first term and the common difference:

$13,53,93,133$ ….

Concept in a Minute:

An Arithmetic Progression (A.P.) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (d). The first term (a) is simply the first number in the sequence.

Explanation:

The given A.P. is $13, 53, 93, 133$ ….

To find the first term (a), we look at the very first number in the sequence.
The first term (a) = 13.

To find the common difference (d), we subtract any term from its succeeding term.
Let’s subtract the first term from the second term:
d = $53 – 13 = 40$

Let’s verify this by subtracting the second term from the third term:
d = $93 – 53 = 40$

And also, by subtracting the third term from the fourth term:
d = $133 – 93 = 40$

Since the difference between consecutive terms is constant, this is indeed an A.P. and the common difference is 40.

Therefore, the first term is 13 and the common difference is 40.

Question:

For the following A.Ps, write the first term and the common difference:

-5, -1, 3, 7

Concept in a Minute:

Explanation:

The given sequence is an arithmetic progression (A.P.): -5, -1, 3, 7.

To find the first term (a), we simply identify the first number in the sequence.
In this case, the first term, a = -5.

To find the common difference (d), we subtract any term from its succeeding term. Let’s check with a few pairs:
First term = -5
Second term = -1
Third term = 3
Fourth term = 7

Common difference (d) = Second term – First term
d = -1 – (-5)
d = -1 + 5
d = 4

Let’s verify with another pair:
Common difference (d) = Third term – Second term
d = 3 – (-1)
d = 3 + 1
d = 4

The common difference is consistently 4.

Therefore, the first term of the A.P. is -5 and the common difference is 4.

Question:

For the following A.Ps, write the first term and the common difference.

0.6, 1.7, 2.8, 3.9

Concept in a Minute:

Explanation:

The given sequence is 0.6, 1.7, 2.8, 3.9.

To find the first term (a), we simply identify the first number in the sequence.
First term (a) = 0.6

To find the common difference (d), we subtract any term from its succeeding term. Let’s take the first two terms:
d = Second term – First term
d = 1.7 – 0.6
d = 1.1

Let’s verify with the next pair of terms:
d = Third term – Second term
d = 2.8 – 1.7
d = 1.1

And the next pair:
d = Fourth term – Third term
d = 3.9 – 2.8
d = 1.1

Since the difference between consecutive terms is constant, the sequence is indeed an Arithmetic Progression.

Therefore, the first term is 0.6 and the common difference is 1.1.

Question:

Write first four terms of the A.P. when the first term a and the common differenced are given as follows:

a = -2, d = 0

Concept in a Minute:

An arithmetic progression (A.P.) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d). The formula for the nth term of an A.P. is a_n = a + (n-1)d, where ‘a’ is the first term and ‘d’ is the common difference. To find the first four terms, we need to calculate a1, a2, a3, and a4.

Explanation:

We are given the first term (a) = -2 and the common difference (d) = 0.
The first term is given as a1 = a = -2.
The second term (a2) is found by adding the common difference to the first term: a2 = a1 + d = -2 + 0 = -2.
The third term (a3) is found by adding the common difference to the second term: a3 = a2 + d = -2 + 0 = -2.
The fourth term (a4) is found by adding the common difference to the third term: a4 = a3 + d = -2 + 0 = -2.
Therefore, the first four terms of the A.P. are -2, -2, -2, -2.

Question:

A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money does the contractor have to pay as a penalty if he has delayed the work by 30 days.

Concept in a Minute:

This problem involves an arithmetic progression (AP). An AP is a sequence of numbers such that the difference between consecutive terms is constant. The sum of an AP can be calculated using a formula.

Explanation:

The penalties for the delay form an arithmetic progression: 200, 250, 300, …
The first term (a) of this AP is 200.
The common difference (d) is 250 – 200 = 50.
The number of terms (n) is the number of days the work is delayed, which is 30.

We need to find the total penalty, which is the sum of the first 30 terms of this AP.
The formula for the sum of the first n terms of an AP is:
Sn = n/2 * [2a + (n-1)d]

Substitute the values:
S30 = 30/2 * [2*200 + (30-1)*50]
S30 = 15 * [400 + (29)*50]
S30 = 15 * [400 + 1450]
S30 = 15 * [1850]
S30 = 27750

Therefore, the contractor has to pay Rs. 27750 as a penalty.

Question:

Fill in the blank in the following table, given that a is the first term, d the common difference, and a_nnth term of the AP:

a	d	n	an
______	-3	18	-5

Concept in a Minute:

The nth term of an arithmetic progression (AP) is given by the formula: a_n = a + (n-1)d, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number.

Explanation:

We are given the values for the common difference (d = -3), the term number (n = 18), and the nth term (a_n = -5). We need to find the first term (a).
Using the formula for the nth term of an AP:
a_n = a + (n-1)d

Substitute the given values into the formula:
-5 = a + (18-1)(-3)

Simplify the expression inside the parenthesis:
-5 = a + (17)(-3)

Perform the multiplication:
-5 = a – 51

To find ‘a’, add 51 to both sides of the equation:
-5 + 51 = a
46 = a

Therefore, the first term (a) is 46.

The filled table:
a = 46
d = -3
n = 18
a_n = -5

Question:

Fill in the blank in the following table, given that a is the first term, d the common difference, and a_n nth term of the AP:

a	d	n	an
-18.9	2.5	______	3.6

Concept in a Minute:

The nth term of an arithmetic progression (AP) is given by the formula: $a_n = a + (n-1)d$, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number. To find a missing value, rearrange the formula to solve for the unknown variable.

Explanation:

We are given the first term ($a = -18.9$), the common difference ($d = 2.5$), and the nth term ($a_n = 3.6$). We need to find the term number ‘n’.
Using the formula $a_n = a + (n-1)d$:
Substitute the given values:
$3.6 = -18.9 + (n-1)2.5$
Add 18.9 to both sides:
$3.6 + 18.9 = (n-1)2.5$
$22.5 = (n-1)2.5$
Divide both sides by 2.5:
$22.5 / 2.5 = n-1$
$9 = n-1$
Add 1 to both sides:
$9 + 1 = n$
$n = 10$
Therefore, the blank should be filled with 10.

Question:

Fill in the blank in the following table, given that a is the first term, d the common difference, and a_n nth term of the AP:

a	d	n	a_n
-18	______	10	0

Concept in a Minute:

The nth term of an arithmetic progression (AP) is given by the formula a_n = a + (n-1)d, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number.

Explanation:

We are given the first term (a = -18), the term number (n = 10), and the nth term (a_n = 0). We need to find the common difference (d).
Using the formula for the nth term of an AP:
a_n = a + (n-1)d
Substitute the given values:
0 = -18 + (10-1)d
0 = -18 + 9d
Now, we need to solve for ‘d’.
Add 18 to both sides of the equation:
18 = 9d
Divide both sides by 9:
d = 18 / 9
d = 2

Therefore, the common difference is 2.

The completed table will be:
a = -18
d = 2
n = 10
a_n = 0

Question:

Fill in the blank in the following table, given that a is the first term, d the common difference, and a_n nth term of the AP:

a	d	n	a_n
7	3	8	______

Concept in a Minute:

The nth term of an arithmetic progression (AP) is given by the formula: a_n = a + (n-1)d, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number.

Explanation:

In this problem, we are given:
The first term, a = 7
The common difference, d = 3
The term number, n = 8

We need to find the 8th term of the AP, which is a_n.
Using the formula for the nth term of an AP:
a_n = a + (n-1)d
Substitute the given values into the formula:
a_8 = 7 + (8-1) * 3
a_8 = 7 + (7) * 3
a_8 = 7 + 21
a_8 = 28

Therefore, the missing value in the table is 28.

The completed table is:
a = 7, d = 3, n = 8, a_n = 28

Question:

Fill in the blank in the following table, given that a is the first term, d the common difference, and a_n nth term of the AP:

a	d	n	an
3.5	0	105	______

Concept in a Minute:

The nth term of an arithmetic progression (AP) is given by the formula: a_n = a + (n-1)d, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number.

Explanation:

We are given the first term (a = 3.5), the common difference (d = 0), and the term number (n = 105). We need to find the nth term (a_n).
Using the formula for the nth term of an AP:
a_n = a + (n-1)d
Substitute the given values into the formula:
a_105 = 3.5 + (105 – 1) * 0
a_105 = 3.5 + (104) * 0
a_105 = 3.5 + 0
a_105 = 3.5

Therefore, the blank should be filled with 3.5.

Next Chapter: Circles

Refer Arithmetic Progressions Notes

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