Class 9 Maths: Sequences and Progressions – Extra Questions

Ans: A

Explanation:
The given sequence is an arithmetic progression (AP) with the first term a = 21 and the common difference d = 18 – 21 = -3.
We want to find the term (n) for which the value is 0.
The nth term of an AP is given by the formula: an = a + (n-1)d
In this case, an = 0, a = 21, and d = -3.
So, 0 = 21 + (n-1)(-3)
-21 = -3(n-1)
7 = n – 1
n = 8

Correct Answer: 8

Ans: C

Explanation:
In an arithmetic progression, the difference between consecutive terms is constant. We can use this fact to find the common difference and the missing values.

Since 7, b, and 23 are consecutive terms, the common difference can be calculated, in part, from the numbers 7 and 23.
The difference between 23 and 7, which spans two terms is 23-7 = 16.
So, to get the common difference, we divide 16 by 2 (since the difference spans two terms) to get 8.

Now we can proceed to find the terms.
Starting from 7, with a common difference of 8:
b = 7 + 8 = 15
To get the term before 7, we subtract the common difference from 7:
a = 7 – 8 = -1
To get the term after 23, we add the common difference to 23:
c = 23 + 8 = 31

Therefore, the arithmetic progression is -1, 7, 15, 23, 31.

Correct Answer: a = -1, b = 15, c = 31

Ans: B

Explanation: This is an arithmetic sequence problem. The first term (a) is 1000, and the common difference (d) is 50. We want to find the 30th term (n=30). The formula for the nth term of an arithmetic sequence is: a_n = a + (n-1)d. So, a_30 = 1000 + (30-1)*50 = 1000 + 29*50 = 1000 + 1450 = 2450.
Correct Option: B

Ans: B

Explanation: Let the first term of the arithmetic progression be $a$ and the common difference be $d$. The $n$th term of an arithmetic progression is given by $a_n = a + (n-1)d$. The 27th term is $a_{27} = a + (27-1)d = a + 26d$. The 7th term is $a_7 = a + (7-1)d = a + 6d$. The difference between the 27th term and the 7th term is $a_{27} – a_7 = (a + 26d) – (a + 6d) = 20d$. We are given that this difference is 80, so $20d = 80$. Dividing both sides by 20, we get $d = 4$.

Correct Option: B

Ans: B

Solution:
The sequence is an arithmetic progression with the first term $a = -7$ and common difference $d = -2 – (-7) = 5$.
The $n$th term is given by $t_n = a + (n-1)d$.
Therefore, $t_n = -7 + (n-1)5 = -7 + 5n – 5 = 5n – 12$.
For $t_{26}$, we have $t_{26} = 5(26) – 12 = 130 – 12 = 118$.
Answer: $t_n = 5n – 12$, $t_{26} = 118$

Ans: B

Explanation: Let the common step size be ‘d’. The first sequence is 3, 3+d, 3+2d, … The 100th term of this sequence is 3 + 99d. The second sequence is 8, 8+d, 8+2d, … The 100th term of this sequence is 8 + 99d. The difference between the 100th terms is (8 + 99d) – (3 + 99d) = 8 – 3 = 5.

Correct Option: B

Ans: B

Solution:
Let the first term be $a$ and the common difference be $d$.
Then the $n$th term of the AP is given by $a_n = a + (n-1)d$.
Given $a_7 = a + 6d = 10$ and $a_{13} = a + 12d = 22$.
Subtracting the first equation from the second equation, we get
$6d = 12$, which implies $d = 2$.
Substituting $d=2$ into the first equation, we get
$a + 6(2) = 10$, which means $a + 12 = 10$, so $a = -2$.
Then $a_{20} = a + 19d = -2 + 19(2) = -2 + 38 = 36$.

Answer: 36

Ans: A

Solution:
The common difference is $(2x+2+y) – (2x+y) = 2$. Also, $(2x+4+y) – (2x+2+y) = 2$. Since the difference is constant, it is an AP.
The next three terms are $(2x+6)+y$, $(2x+8)+y$, and $(2x+10)+y$.

Answer: $(2x+6)+y$, $(2x+8)+y$, $(2x+10)+y$

Ans: D

Explanation:
The given arithmetic progression is 18, 15.5, 13, …, -47.
The first term, a = 18.
The common difference, d = 15.5 – 18 = -2.5.
The last term, l = -47.
We use the formula for the nth term of an arithmetic progression:
l = a + (n-1)d
Substituting the given values:
-47 = 18 + (n-1)(-2.5)
-47 – 18 = (n-1)(-2.5)
-65 = (n-1)(-2.5)
n-1 = -65 / -2.5
n-1 = 26
n = 26 + 1
n = 27

Correct Answer: 27

Ans: A

Solution:
The given sequence is an arithmetic progression with first term $a_1 = -4$ and common difference $d = -2 – (-4) = 2$.
The sum of an arithmetic series is given by $S_n = \frac{n}{2}(a_1 + a_n)$.
We are given that $S_n = 432$, $a_1 = -4$, and $a_n = x$.
We also have $a_n = a_1 + (n-1)d$, so $x = -4 + (n-1)(2)$.
Thus, $432 = \frac{n}{2}(-4 + x)$.
Substituting $x = -4 + 2(n-1)$ into the sum formula:
$432 = \frac{n}{2}(-4 -4 + 2n – 2) = \frac{n}{2}(2n – 10) = n^2 – 5n$
$n^2 – 5n – 432 = 0$
$(n – 24)(n + 18) = 0$
Since $n$ must be positive, $n = 24$.
Then $x = -4 + (24-1)(2) = -4 + 23(2) = -4 + 46 = 42$.

Answer: 42

Ans: C

Solution:
1. Calculate John’s total savings: 1 + 2 + 3 + … + 31 = (31 * 32) / 2 = 496
2. Calculate John’s total allowance: Savings + Spending + Leftover = 496 + 150 + 50 = 696

Answer: 696

Ans: A

Explanation: Let the common difference be ‘d’. The first four terms are 5, 5+d, 5+2d, and 5+3d. The sum of these terms is 5 + (5+d) + (5+2d) + (5+3d) = 20 + 6d. The next four terms are 5+4d, 5+5d, 5+6d, and 5+7d. The sum of these terms is (5+4d) + (5+5d) + (5+6d) + (5+7d) = 20 + 22d. According to the problem, the sum of the first four terms is half the sum of the subsequent four terms, so 20 + 6d = (1/2)(20 + 22d). Multiplying both sides by 2 gives 40 + 12d = 20 + 22d. Subtracting 12d from both sides gives 40 = 20 + 10d. Subtracting 20 from both sides gives 20 = 10d. Dividing by 10 gives d = 2.

Correct Answer: 2

Ans: C

Solution:
Let the first term be $a$ and the common difference be $d$.
Then $a_n = a + (n-1)d$.
We are given that $a_4 + a_9 = 10$ and $a_8 + a_{15} = -4$.
So $(a + 3d) + (a + 8d) = 10$, which means $2a + 11d = 10$.
Also, $(a + 7d) + (a + 14d) = -4$, which means $2a + 21d = -4$.
Subtracting the first equation from the second equation gives:
$(2a + 21d) – (2a + 11d) = -4 – 10$
$10d = -14$, so $d = -1.4$.
Substituting $d = -1.4$ into $2a + 11d = 10$:
$2a + 11(-1.4) = 10$
$2a – 15.4 = 10$
$2a = 25.4$
$a = 12.7$
We want to find the 11th term, $a_{11} = a + 10d = 12.7 + 10(-1.4) = 12.7 – 14 = -1.3$

Answer: -1.3

Ans: A

Explanation:
To verify that the given sequence is an arithmetic progression (AP), we need to check if the difference between consecutive terms is constant. Let’s find the difference between the second and first terms: $2\sqrt{3} – \sqrt{3} = \sqrt{3}$. Now, find the difference between the third and second terms: $3\sqrt{3} – 2\sqrt{3} = \sqrt{3}$. Since the difference between consecutive terms is constant ($\sqrt{3}$), the sequence is an AP. The common difference, d, is $\sqrt{3}$.

To find the next three terms, we add the common difference to the last given term.
The fourth term: $3\sqrt{3} + \sqrt{3} = 4\sqrt{3}$
The fifth term: $4\sqrt{3} + \sqrt{3} = 5\sqrt{3}$
The sixth term: $5\sqrt{3} + \sqrt{3} = 6\sqrt{3}$

Correct Answer: $4\sqrt{3}, 5\sqrt{3}, 6\sqrt{3}$

Ans: D

The given sequence is a harmonic progression (HP). The reciprocals of the terms form an arithmetic progression (AP).

The reciprocals are: $5, \frac{19}{4}, \frac{9}{2}, \frac{17}{4}, …$

The reciprocals of the given sequence terms are NOT forming AP. The sequence provided is NOT a Harmonic Progression.

Let’s assume there is a typo in the given sequence. Suppose the sequence is $\frac{1}{5}, \frac{1}{9}, \frac{1}{13}, \frac{1}{17},…$ which implies the reciprocals are $5, 9, 13, 17,…$ which is an AP. The common difference of this AP is 4. The 10th term of the AP will be $a_n = a_1 + (n-1)d$, i.e. $5 + (10-1)4 = 5 + 36 = 41$.

Therefore, the 10th term of the HP is $\frac{1}{41}$.

Ans: B

Solution:
Let the common difference be d.
Sum of first four terms: S4 = 4/2 * [2*3 + (4-1)*d] = 2(6+3d) = 12+6d
Sum of next four terms: S8 – S4 = 8/2 * [2*3 + (8-1)*d] – (12+6d) = 4(6+7d) – 12 – 6d = 24+28d-12-6d = 12+22d
Given: 12+6d = 1/3 * (12+22d)
36+18d = 12+22d
24 = 4d
d=6
a15 = 3 + (15-1)*6 = 3+14*6 = 3+84 = 87

Answer: 87

Ans: B

Solution:
Let $S_k$ denote the sum of the first $k$ terms.
We have $S_m = n$ and $S_n = m$.
The general form for the sum of an arithmetic progression is $S_k = \frac{k}{2}[2a + (k-1)d]$, where $a$ is the first term and $d$ is the common difference.
Thus, $\frac{m}{2}[2a + (m-1)d] = n$ and $\frac{n}{2}[2a + (n-1)d] = m$.
$2am + m(m-1)d = 2n$ and $2an + n(n-1)d = 2m$.
Subtract the two equations multiplied by appropriate factors, $2a(m-n) + (m^2-m-n^2+n)d = 2(n-m)$.
$2a(m-n) + (m-n)(m+n)d – (m-n)d = 2(n-m)$.
Since $m \neq n$, $(m-n)(2a+(m+n-1)d) = 2(n-m)$.
$2a+(m+n-1)d = \frac{2(n-m)}{m-n} = -2$.
Thus $S_{m+n} = \frac{m+n}{2}[2a + (m+n-1)d] = \frac{m+n}{2}(-2) = -(m+n)$.
Answer: $-(m+n)$

Ans: D

Solution:
The given sequence is an arithmetic progression with first term $a_1 = -4$ and common difference $d = -1 – (-4) = 3$.
Let $n$ be the number of terms. Then $a_n = -4 + (n-1)3 = 3n – 7$.
The sum of the first $n$ terms is $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(-4 + 3n – 7) = \frac{n(3n – 11)}{2}$.
We are given that $S_n = 440$, so $\frac{n(3n – 11)}{2} = 440$, which gives $n(3n – 11) = 880$.
Thus, $3n^2 – 11n – 880 = 0$. Factoring the quadratic, we have $(3n + 55)(n – 16) = 0$.
Since $n$ must be a positive integer, we have $n = 16$.
Therefore, $x = a_{16} = 3(16) – 7 = 48 – 7 = 41$.

Answer: 41

Ans: C

Explanation: Let $a$ be the first term and $d$ be the common difference of the arithmetic progression. The sum of the first $n$ terms of an AP is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
We are given that $S_{16} = 1624$, so we have
$ \frac{16}{2}[2a + (16-1)d] = 1624 \\ 8[2a + 15d] = 1624 \\ 2a + 15d = \frac{1624}{8} \\ 2a + 15d = 203 $
We are also given that the first term is 500 times the common difference, which means $a = 500d$. Substituting this into the equation $2a + 15d = 203$:
$ 2(500d) + 15d = 203 \\ 1000d + 15d = 203 \\ 1015d = 203 \\ d = \frac{203}{1015} \\ d = \frac{1}{5} $
Thus, the common difference is $\frac{1}{5}$.

Correct Option: C

Ans: C

Solution:
Let the corresponding arithmetic progression (AP) be $a_n$.
Since the 2nd term of the HP is $\frac{1}{7}$, the 2nd term of the AP is 7.
Since the 5th term of the HP is $\frac{1}{16}$, the 5th term of the AP is 16.
Let the first term of the AP be $a$ and the common difference be $d$.
$a+d = 7$
$a+4d = 16$
Subtracting the first equation from the second, we get $3d=9$, so $d=3$.
Substituting $d=3$ into the first equation, $a+3=7$, so $a=4$.
The 8th term of the AP is $a+7d = 4+7(3) = 4+21=25$.
The 8th term of the HP is $\frac{1}{25}$.

Answer: $\frac{1}{25}$