SEBA Class 10 Maths Chapter 5: Arithmetic Progressions Miscellaneous Questions Solutions
Get free, step-by-step solutions to SEBA Class 10 Maths Chapter 5: Arithmetic Progressions Miscellaneous Questions and MCQs from the new SCERT textbook 2026. In this article, we solve Miscellaneous Questions 50 to 65—including MCQs, Very Short Answer (VSA) questions, and Short Answer (SA) questions—in a clear, systematic manner.
See More:
- SEBA Class 10 Maths Exercise 5.1 Solutions: Arithmetic Progressions
- SEBA Class 10 Maths Exercise 5.2 Solutions: Arithmetic Progressions
- SEBA Class 10 Maths Exercise 5.3 Solutions: Arithmetic Progressions
This Section Comprises Multiple Choice Questions (MCQs)
50. The common difference of the A.P. whose $n$th term is given by $a_n = 5n – 7$ is:
(a) -7 (b) 7 (c) 5 (d) -2
$\mathbf{Sol^n.}$
Given:
$a_n = 5n – 7$
Putting $n = 1$:
$a_1 = 5(1) – 7 = 5 – 7 = -2$
Putting $n = 2$:
$a_2 = 5(2) – 7 = 10 – 7 = 3$
Now, common difference:
$d = a_2 – a_1 = 3 – (-2) = 3 + 2 = 5$
Correct option is (c) 5
51. The 13th term from the end of the A.P.: $20, 13, 6, -1, \dots, -148$ is:
(a) 57 (b) -57 (c) 64 (d) -64
$\mathbf{Sol^n.}$
Given AP: $20, 13, 6, -1, \dots, -148$
Writing the AP in reverse order:
$-148, \dots, -1, 6, 13, 20$
Here:
First term, $a = -148$
Common difference, $d = 20 – 13 = 7$
Number of terms, $n = 13$
We know that:
$a_n = a + (n – 1)d$
$\implies a_{13} = -148 + (13 – 1)(7)$
$= -148 + 12 \times 7$
$= -148 + 84$
$= -64$
Correct option is (d) -64
52. The $11^{\text{th}}$ term from the end of the A.P.: $10, 7, 4, \dots, -62$ is:
(a) 25 (b) 16 (c) -32 (d) 0
$\mathbf{Sol^n.}$
Given AP: $10, 7, 4, \dots, -62$
Writing the AP in reverse order:
$-62, \dots, 4, 7, 10$
Here:
First term, $a = -62$
Common difference, $d = 10 – 7 = 3$
Number of terms, $n = 11$
We know that:
$a_n = a + (n – 1)d$
$\implies a_{11} = -62 + (11 – 1)(3)$
$= -62 + 10 \times 3$
$= -62 + 30$
$= -32$
Correct option is (c) -32
53. The common difference of the A.P. whose $n$th term is given by $a_n = 3n + 7$ is:
(a) 7 (b) 3 (c) $3n$ (d) 1
$\mathbf{Sol^n.}$
Given:
$a_n = 3n + 7$
Putting $n = 1$:
$a_1 = 3(1) + 7 = 3 + 7 = 10$
Putting $n = 2$:
$a_2 = 3(2) + 7 = 6 + 7 = 13$
Now, common difference:
$d = a_2 – a_1 = 13 – 10 = 3$
Correct option is (b) 3
54. The 8th term of an A.P. is 17 and its 14th term is 29. The common difference of this A.P. is:
(a) 3 (b) 2 (c) 5 (d) -2
$\mathbf{Sol^n.}$
Given:
$a_8 = 17 \implies a + 7d = 17 \quad \text{— (1)}$
$a_{14} = 29 \implies a + 13d = 29 \quad \text{— (2)}$
Subtracting equation (1) from equation (2):
$(a + 13d) – (a + 7d) = 29 – 17$
$\implies 6d = 12$
$\implies d = \frac{12}{6}$
$\implies d = 2$
Correct option is (b) 2
55. The sum of the first 100 even natural numbers is:
(a) 10100 (b) 2550 (c) 5050 (d) 10010
$\mathbf{Sol^n.}$ The first 100 even natural numbers are: $2, 4, 6, 8, \dots$ Here, $a = 2$, $d = 4 – 2 = 2$, and $n = 100$
We know that:
$S_n = \frac{n}{2}[2a + (n – 1)d]$
$\implies S_{100} = \frac{100}{2}[2(2) + (100 – 1)2]$
$= 50[4 + 99 \times 2]$
$= 50[4 + 198]$
$= 50 \times 202$
$= 10100$
(Alternatively, sum of first $n$ even natural numbers is $n(n + 1) = 100 \times 101 = 10100$)
Correct option is (a) 10100
56. The sum of the first 50 odd natural numbers is:
(a) 5000 (b) 2500 (c) 2550 (d) 5050
$\mathbf{Sol^n.}$
The first 50 odd natural numbers are:
$1, 3, 5, 7, \dots$
The above sequence forms an AP where:
First term, $a = 1$
Common difference, $d = 3 – 1 = 2$
Number of terms, $n = 50$
We know that:
$S_n = \frac{n}{2}[2a + (n – 1)d]$
$\implies S_{50} = \frac{50}{2}[2(1) + (50 – 1)2]$
$= 25[2 + 49 \times 2]$
$= 25[2 + 98]$
$= 25 \times 100$
$= 2500$
$\therefore$ The sum of the first 50 odd natural numbers is 2500.
Correct option is (b) 2500
Alternatively:
We know that the sum of the first $n$ odd natural numbers is given by: $S_n = n^2$
Here, $n = 50$
$S_{50} = (50)^2$
$= 2500$
Correct option is (b) 2500
57. The $4^{\text{th}}$ term from the end of the A.P.: $-8, -5, -2, \dots, 49$ is:
(a) 37 (b) 40 (c) 1 (d) 43
$\mathbf{Sol^n.}$
Given AP: $-8, -5, -2, \dots, 49$
Writing the AP in reverse order:
$49, \dots, -2, -5, -8$
Here:
First term, $a = 49$
Common difference, $d = -5 – (-2) = -5 + 2 = -3$
Number of terms, $n = 4$
We know that:
$a_n = a + (n – 1)d$
$\implies a_4 = 49 + (4 – 1)(-3)$
$= 49 + 3(-3)$
$= 49 – 9$
$= 40$
Correct option is (b) 40
58. If $n$th term of an A.P. is $7n – 4$, then the common difference of the A.P. is:
(a) 7 (b) $7n$ (c) -4 (d) 4
$\mathbf{Sol^n.}$
Given:
$a_n = 7n – 4$
Putting $n = 1$:
$a_1 = 7(1) – 4 = 7 – 4 = 3$
Putting $n = 2$:
$a_2 = 7(2) – 4 = 14 – 4 = 10$
Now, common difference:
$d = a_2 – a_1 = 10 – 3 = 7$
Correct option is (a) 7
This Section Comprises Very Short Answer (VSA) Type Questions
59. Find the sum of the first 20 terms of the A.P.: $1/15, 1/12, 1/10, \dots$
$\mathbf{Sol^n.}$
Given AP: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$
Here, $a = \frac{1}{15}$ and $n = 20$
$d = \frac{1}{12} – \frac{1}{15} = \frac{5 – 4}{60} = \frac{1}{60}$
We know that:
$S_n = \frac{n}{2}[2a + (n – 1)d]$
$\implies S_{20} = \frac{20}{2}\left[2\left(\frac{1}{15}\right) + (20 – 1)\left(\frac{1}{60}\right)\right]$
$= 10\left[\frac{2}{15} + 19 \times \frac{1}{60}\right]$
$= 10\left[\frac{2}{15} + \frac{19}{60}\right]$
$= 10\left[\frac{8 + 19}{60}\right]$
$= 10 \times \frac{27}{60}$
$= \frac{27}{6}$
$= \frac{9}{2}$
$\therefore$ The sum of the first 20 terms of the given AP is $\frac{9}{2}$ (or $4\frac{1}{2}$).
60. Find the sum of the first 15 terms of the A.P.: $2/3, 0, -2/3, -4/3, \dots$
$\mathbf{Sol^n.}$
Given AP: $\frac{2}{3}, 0, -\frac{2}{3}, -\frac{4}{3}, \dots$
Here, $a = \frac{2}{3}$ and $n = 15$
$d = 0 – \frac{2}{3} = -\frac{2}{3}$
We know that:
$S_n = \frac{n}{2}[2a + (n – 1)d]$
$\implies S_{15} = \frac{15}{2}\left[2\left(\frac{2}{3}\right) + (15 – 1)\left(-\frac{2}{3}\right)\right]$
$= \frac{15}{2}\left[\frac{4}{3} + 14\left(-\frac{2}{3}\right)\right]$
$= \frac{15}{2}\left[\frac{4}{3} – \frac{28}{3}\right]$
$= \frac{15}{2}\left[-\frac{24}{3}\right]$
$= \frac{15}{2} \times (-8)$
$= 15 \times (-4)$
$= -60$
$\therefore$ The sum of the first 15 terms of the given AP is -60.
61. Find the sum of the first 20 terms of the A.P.: $-29/3, -9, -25/3, -23/3, \dots$
$\mathbf{Sol^n.}$
Given AP: $-\frac{29}{3}, -9, -\frac{25}{3}, -\frac{23}{3}, \dots$
Here, $a = -\frac{29}{3}$ and $n = 20$
$d = -9 – \left(-\frac{29}{3}\right) = -9 + \frac{29}{3} = \frac{-27 + 29}{3} = \frac{2}{3}$
We know that:
$S_n = \frac{n}{2}[2a + (n – 1)d]$
$\implies S_{20} = \frac{20}{2}\left[2\left(-\frac{29}{3}\right) + (20 – 1)\left(\frac{2}{3}\right)\right]$
$= 10\left[-\frac{58}{3} + 19\left(\frac{2}{3}\right)\right]$
$= 10\left[-\frac{58}{3} + \frac{38}{3}\right]$
$= 10\left[\frac{-58 + 38}{3}\right]$
$= 10\left[-\frac{20}{3}\right]$
$= -\frac{200}{3}$
$\therefore$ The sum of the first 20 terms of the given AP is $-\frac{200}{3}$ (or $-66\frac{2}{3}$).
62. (a) The sum of the first 12 terms of an A.P is 900. If its first term is 20, then find the common difference and 12th term.
$\mathbf{Sol^n.}$
Given:
$n = 12$, $a = 20$, and $S_{12} = 900$
We know that:
$S_n = \frac{n}{2}[2a + (n – 1)d]$
$\implies 900 = \frac{12}{2}[2(20) + (12 – 1)d]$
$\implies 900 = 6[40 + 11d]$
$\implies \frac{900}{6} = 40 + 11d$
$\implies 150 = 40 + 11d$
$\implies 11d = 150 – 40$
$\implies 11d = 110$
$\implies d = \frac{110}{11}$
$\implies d = 10$
Now, finding the 12th term ($a_{12}$):
$a_{12} = a + 11d$
$= 20 + 11(10)$
$= 20 + 110$
$= 130$
$\therefore$ The common difference is 10 and the 12th term is 130.
OR
62. (b) The sum of first $n$ terms of an A.P. is represented by $S_n = 6n – n^2$. Find the common difference.
$\mathbf{Sol^n.}$
Given:
$S_n = 6n – n^2$
Putting $n = 1$:
$S_1 = 6(1) – (1)^2 = 6 – 1 = 5$
$\therefore a_1 = a = S_1 = 5$
Putting $n = 2$:
$S_2 = 6(2) – (2)^2 = 12 – 4 = 8$
Now, second term:
$a_2 = S_2 – S_1$
$= 8 – 5$
$= 3$
Now, common difference:
$d = a_2 – a_1$
$= 3 – 5$
$= -2$
$\therefore$ The common difference is -2.
This Section Comprises Short Answer (SA) Type Questions
63. In an A.P., the sum of the first $n$ terms is given by $S_n = 6n – n^2$. Find its 30th term.
$\mathbf{Sol^n.}$
Given:
$S_n = 6n – n^2$
Putting $n = 1$:
$S_1 = 6(1) – (1)^2 = 6 – 1 = 5$
$\therefore a_1 = a = S_1 = 5$
Putting $n = 2$:
$S_2 = 6(2) – (2)^2 = 12 – 4 = 8$
Now, second term:
$a_2 = S_2 – S_1 = 8 – 5 = 3$
Common difference:
$d = a_2 – a_1 = 3 – 5 = -2$
Now, finding the 30th term ($a_{30}$):
$a_{30} = a + (30 – 1)d$
$= a + 29d$
$= 5 + 29(-2)$
$= 5 – 58$
$= -53$
$\therefore$ The 30th term of the AP is -53.
64. Find the common difference of an A.P. whose first term is 8, the last term is 65 and sum of all terms is 730.
$\mathbf{Sol^n.}$
Given:
First term, $a = 8$
Last term, $l = a_n = 65$
Sum of terms, $S_n = 730$
We know that:
$S_n = \frac{n}{2}(a + l)$
$\implies 730 = \frac{n}{2}(8 + 65)$
$\implies 730 = \frac{n}{2}(73)$
$\implies \frac{730 \times 2}{73} = n$
$\implies 10 \times 2 = n$
$\implies n = 20$
Now, using $a_n = a + (n – 1)d$:
$a_{20} = a + 19d = 65$
$\implies 8 + 19d = 65$
$\implies 19d = 65 – 8$
$\implies 19d = 57$
$\implies d = \frac{57}{19}$
$\implies d = 3$
$\therefore$ The common difference of the AP is 3.
65. If $p$th term of an A.P. is $q$ and $q$th term is $p$, then prove that its $n$th term is $(p + q – n)$.
$\mathbf{Sol^n.}$
Let the first term of the AP be $a$ and the common difference be $d$.
According to the question:
$a_p = q \implies a + (p – 1)d = q \quad \text{— (1)}$
$a_q = p \implies a + (q – 1)d = p \quad \text{— (2)}$
Subtracting equation (2) from equation (1):
$[a + (p – 1)d] – [a + (q – 1)d] = q – p$
$\implies a + pd – d – a – qd + d = q – p$
$\implies (p – q)d = -(p – q)$
$\implies d = \frac{-(p – q)}{p – q}$
$\implies d = -1$
Substituting $d = -1$ in equation (1):
$a + (p – 1)(-1) = q$
$\implies a – p + 1 = q$
$\implies a = p + q – 1 \quad \text{— (3)}$
Now, finding the $n$th term ($a_n$):
$a_n = a + (n – 1)d$
Substituting the values of $a$ and $d$:
$a_n = (p + q – 1) + (n – 1)(-1)$
$= p + q – 1 – n + 1$
$= p + q – n$
$\therefore a_n = (p + q – n)$
Hence proved.