Class 10 Maths: Arithmetic Progressions – Extra Questions with Answers
Q. 1 Determine if the following sequence is an arithmetic progression (AP). If it is, find the first term, common difference, and the next term in the series: $2, \frac{5}{2}, 3, \frac{7}{2}, …$
Check Solution
Ans: C
Solution:
The sequence is an AP.
First term (a) = 2
Common difference (d) = $\frac{5}{2} – 2 = \frac{1}{2}$
Next term = $\frac{7}{2} + \frac{1}{2} = 4$
Answer: a = 2, d = 1/2, next term = 4
Q. 2 If $x-2$, $6$, $x+2$ are in AP, find the value of $x$.
Check Solution
Ans: A
Solution:
Since $x-2$, $6$, $x+2$ are in AP, the difference between consecutive terms is constant.
Therefore, $6 – (x-2) = (x+2) – 6$.
Simplifying, $6-x+2 = x+2-6$.
$8-x = x-4$.
$2x = 12$.
$x=6$.
Answer: 6
Q. 3 Evaluate $\sum_{r=1}^{5} (4 + 3^r)$.
Check Solution
Ans: A
Solution:
$\sum_{r=1}^{5} (4 + 3^r) = \sum_{r=1}^{5} 4 + \sum_{r=1}^{5} 3^r = 4 \cdot 5 + (3^1 + 3^2 + 3^3 + 3^4 + 3^5) = 20 + (3 + 9 + 27 + 81 + 243) = 20 + 363 = 383$
Answer: 383
Q. 4 If the sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 2n$, find the first three terms of the AP.
Check Solution
Ans: B
Solution:
$S_1 = 3(1)^2 + 2(1) = 5 = a_1$
$S_2 = 3(2)^2 + 2(2) = 16$
$a_2 = S_2 – S_1 = 16 – 5 = 11$
$d = a_2 – a_1 = 11 – 5 = 6$
$a_3 = a_2 + d = 11 + 6 = 17$
Answer: 5, 11, 17
Q. 5 A concert venue charges a penalty for exceeding the planned concert duration. The penalty is Rs 1000 for the first 30 minutes, Rs 1100 for the second 30 minutes, Rs 1200 for the third 30 minutes, and so on, with each subsequent 30-minute increment costing Rs 100 more. If the concert runs for an extra 2 hours and 30 minutes, what is the total penalty the venue must pay?
Check Solution
Ans: A
Solution:
2 hours and 30 minutes is equal to 5 increments of 30 minutes. The penalty amounts form an arithmetic progression: 1000, 1100, 1200, 1300, 1400. The sum of an arithmetic series is given by: S = (n/2) * (2a + (n-1)d) where n is the number of terms, a is the first term, and d is the common difference. In this case, n=5, a=1000, and d=100. S = (5/2) * (2*1000 + (5-1)*100) = (5/2) * (2000 + 400) = (5/2) * 2400 = 6000.
Answer: 6000
Q. 6 The value of $n$ for which the $n^{th}$ terms of the A.P.s $5, 13, 21, \dots$ and $40, 43, 46, \dots$ are equal.
Check Solution
Ans: A
Solution:
Let $a_n$ be the $n$th term of the first A.P. and $b_n$ be the $n$th term of the second A.P.
For the first A.P., the first term is $a_1 = 5$ and the common difference is $d_1 = 13 – 5 = 8$. So, $a_n = a_1 + (n-1)d_1 = 5 + (n-1)8 = 5 + 8n – 8 = 8n – 3$.
For the second A.P., the first term is $b_1 = 40$ and the common difference is $d_2 = 43 – 40 = 3$. So, $b_n = b_1 + (n-1)d_2 = 40 + (n-1)3 = 40 + 3n – 3 = 3n + 37$.
We want to find $n$ such that $a_n = b_n$.
Therefore, $8n – 3 = 3n + 37$.
$8n – 3n = 37 + 3$
$5n = 40$
$n = 8$
Answer: 8
Q. 7 Let $a, b, c, d,$ and $e$ be five real numbers such that $a, b, c$ are in AP, $b, c, d$ are in GP, and $c, d, e$ are in HP. If $a = 3$ and $e = 12$, then find the values of $b, c,$ and $d$.
Check Solution
Ans: A
Solution:
Since $a, b, c$ are in AP, $2b = a+c$.
Since $b, c, d$ are in GP, $c^2 = bd$.
Since $c, d, e$ are in HP, $\frac{2}{d} = \frac{1}{c} + \frac{1}{e}$.
Given $a=3$ and $e=12$, we have $2b = 3+c$, $c^2 = bd$, and $\frac{2}{d} = \frac{1}{c} + \frac{1}{12}$.
From $2b = 3+c$, we get $b = \frac{3+c}{2}$.
From $\frac{2}{d} = \frac{1}{c} + \frac{1}{12} = \frac{12+c}{12c}$, we have $d = \frac{24c}{12+c}$.
Substituting $b$ and $d$ into $c^2 = bd$, we have $c^2 = \frac{3+c}{2} \cdot \frac{24c}{12+c}$.
So $c^2 = \frac{12c(3+c)}{12+c}$.
Since $c \neq 0$, we can divide by $c$: $c = \frac{12(3+c)}{12+c}$.
$c(12+c) = 36 + 12c$
$12c + c^2 = 36 + 12c$
$c^2 = 36$
$c = \pm 6$.
If $c=6$, then $b=\frac{3+6}{2} = \frac{9}{2}$ and $d = \frac{24 \cdot 6}{12+6} = \frac{144}{18} = 8$.
If $c=-6$, then $b=\frac{3-6}{2} = -\frac{3}{2}$ and $d = \frac{24 \cdot (-6)}{12-6} = \frac{-144}{6} = -24$.
When $c=6$, $b, c, d$ are in GP: $\frac{9}{2}, 6, 8$.
$\frac{6}{9/2} = \frac{12}{9} = \frac{4}{3}$ and $\frac{8}{6} = \frac{4}{3}$.
$a, b, c$ are in AP: $3, \frac{9}{2}, 6$. $\frac{9}{2}-3 = \frac{3}{2}$ and $6-\frac{9}{2} = \frac{3}{2}$.
When $c=-6$, $b, c, d$ are in GP: $-\frac{3}{2}, -6, -24$.
$\frac{-6}{-3/2} = 4$ and $\frac{-24}{-6} = 4$.
$a, b, c$ are in AP: $3, -\frac{3}{2}, -6$.
$-\frac{3}{2} – 3 = -\frac{9}{2}$ and $-6-(-\frac{3}{2}) = -6+\frac{3}{2} = -\frac{9}{2}$.
Since $a, b, c$ are in AP and $b, c, d$ are in GP and $c, d, e$ are in HP, and $e=12$, we can conclude that $c>0$, so we choose $c=6$.
Answer: $b=\frac{9}{2}, c=6, d=8$
Q. 8 How many terms of the A.P. 5, 11, 17, … must be taken so that their sum is 455?
Check Solution
Ans: C
Solution:
Let $n$ be the number of terms.
The first term $a = 5$, and the common difference $d = 11 – 5 = 6$.
The sum of an A.P. is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
We are given $S_n = 455$.
So, $455 = \frac{n}{2}[2(5) + (n-1)6]$
$910 = n[10 + 6n – 6]$
$910 = n(4 + 6n)$
$910 = 4n + 6n^2$
$6n^2 + 4n – 910 = 0$
$3n^2 + 2n – 455 = 0$
Using the quadratic formula,
$n = \frac{-2 \pm \sqrt{2^2 – 4(3)(-455)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 5460}}{6} = \frac{-2 \pm \sqrt{5464}}{6} = \frac{-2 \pm 74}{6}$
Since $n$ must be positive, $n = \frac{-2 + 74}{6} = \frac{72}{6} = 12$.
Answer: 12
Q. 9 If $x_1, x_2, x_3, \dots$ are in A.P. and $x_1 + x_7 + x_{10} = -9$ and $x_3 + x_8 + x_{12} = -16$, then $x_3 + x_8 + x_{22} = ?$
Check Solution
Ans: B
Solution:
Since $x_1, x_2, x_3, \dots$ are in A.P., let the common difference be $d$.
We have $x_1 + x_7 + x_{10} = x_1 + (x_1 + 6d) + (x_1 + 9d) = 3x_1 + 15d = -9$.
Dividing by 3, we get $x_1 + 5d = -3$, which means $x_6 = -3$.
We also have $x_3 + x_8 + x_{12} = (x_1 + 2d) + (x_1 + 7d) + (x_1 + 11d) = 3x_1 + 20d = -16$.
Now, we want to find $x_3 + x_8 + x_{22} = (x_1 + 2d) + (x_1 + 7d) + (x_1 + 21d) = 3x_1 + 30d = 3(x_1 + 10d)$.
We have $3x_1 + 15d = -9$, and $3x_1 + 20d = -16$. Subtracting these, we get $5d = -7$, so $d = -\frac{7}{5}$.
Then $x_1 + 5d = x_1 + 5(-\frac{7}{5}) = x_1 – 7 = -3$, so $x_1 = 4$.
Now $x_3 + x_8 + x_{22} = 3x_1 + 30d = 3(4) + 30(-\frac{7}{5}) = 12 – 42 = -30$.
Alternatively, we know that $3x_1 + 15d = -9$ and $3x_1 + 20d = -16$.
Subtracting, we get $-5d = 7$, so $d = -7/5$.
Now we calculate $3x_1 + 30d = 3x_1 + 15d + 15d = -9 + 15d = -9 + 15(-7/5) = -9 – 21 = -30$.
Thus $x_3 + x_8 + x_{22} = 3x_1 + 30d = -30$.
Answer: -30
Q. 10 The seventh term of an arithmetic progression (AP) is double its third term, and the tenth term exceeds one-fourth of its fifth term by 2. Find the 13th term.
Check Solution
Ans: B
Solution:
Let the first term be ‘a’ and the common difference be ‘d’.
7th term: a + 6d
3rd term: a + 2d
Given: a + 6d = 2(a + 2d) => a + 6d = 2a + 4d => a – 2d = 0 … (1)
10th term: a + 9d
5th term: a + 4d
Given: a + 9d = (1/4)(a + 4d) + 2 => 4a + 36d = a + 4d + 8 => 3a + 32d = 8 … (2)
From (1), a = 2d. Substituting in (2): 3(2d) + 32d = 8 => 6d + 32d = 8 => 38d = 8 => d = 4/19
a = 2d = 8/19
13th term: a + 12d = 8/19 + 12(4/19) = 8/19 + 48/19 = 56/19
Answer: 56/19
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