What is an Arithmetic Progression?

An Arithmetic Progression is a sequence where the difference between consecutive terms remains constant.

What is the nth term formula?

The nth term formula is aₙ = a + (n−1)d.

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It provides step-by-step solutions and exam-oriented preparation.

Where can I find Exercise 5.2 solutions?

You can find complete solutions on Digital Pipal Academy.

Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 Solutions | English Medium

Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 Solutions | English Medium | SEBA/SCERT Assam 2026–2027

Are you searching for the best Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 Solutions in English Medium? Here you will get complete and step-by-step solutions based on the latest SEBA/SCERT Assam 2026–2027 syllabus.

Exercise 5.2 is one of the most important parts of Arithmetic Progression (AP). In this exercise, students learn how to find the nth term of an AP using formulas and pattern-based calculations.

Get complete Class 10 Maths Chapter 5 Exercise 5.2 solutions in English Medium based on SEBA/SCERT Assam 2026–2027 syllabus. Learn Arithmetic Progressions and nth term problems step-by-step with easy explanations.

Exercise 5.2 of Class 10 Maths Chapter 5 teaches students how to find the nth term of an arithmetic progression using formulas step by step.

1. Fill in the blanks in the following table, given that $a$ is the first term, $d$ is the common difference and $a_n$ the $n^{th}$ term of the AP:

Step-by-Step Solutions:

We know that the formula for the $n^{th}$ term of an Arithmetic Progression is: $a_n = a + (n - 1)d$

(i) Given: $a = 7, d = 3, n = 8$
$a_n = a + (n - 1)d$
$a_n = 7 + (8 - 1)3 = 7 + (7)3 = 7 + 21 = \mathbf{28}$

(ii) Given: $a = -18, n = 10, a_n = 0$
$a_n = a + (n - 1)d$
$0 = -18 + (10 - 1)d$
$18 = 9d \Rightarrow d = \frac{18}{9} = \mathbf{2}$

(iii) Given: $d = -3, n = 18, a_n = -5$
$a_n = a + (n - 1)d$
$-5 = a + (18 - 1)(-3)$
$-5 = a + 17(-3) \Rightarrow -5 = a - 51$
$a = -5 + 51 = \mathbf{46}$

(iv) Given: $a = -18.9, d = 2.5, a_n = 3.6$
$a_n = a + (n - 1)d$
$3.6 = -18.9 + (n - 1)2.5$
$3.6 + 18.9 = (n - 1)2.5 \Rightarrow 22.5 = (n - 1)2.5$
$n - 1 = \frac{22.5}{2.5} = 9 \Rightarrow n = 9 + 1 = \mathbf{10}$

(v) Given: $a = 3.5, d = 0, n = 105$
$a_n = a + (n - 1)d$
$a_n = 3.5 + (105 - 1)0$
$a_n = 3.5 + 0 = \mathbf{3.5}$

2. Choose the correct choice in the following and justify:

(i) $30^{th}$ term of the AP: $10,\; 7,\; 4,\; \ldots$, is—

(A) $97$ (B) $77$ (C) $\mathbf{-77}$ (D) $-87$

Solution:
Given AP: $10, 7, 4, \dots$
Here, First term ($a$) = $10$
Common difference ($d$) = $7 - 10 = -3$
Number of terms ($n$) = $30$
We know that, $a_n = a + (n - 1)d$
$\therefore a_{30} = 10 + (30 - 1)(-3)$
$= 10 + 29 \times (-3)$
$= 10 - 87 = \mathbf{-77}$
Hence, the correct choice is (C).

(ii) $11^{th}$ term of the AP: $-3,\; -\frac{1}{2},\; 2,\; \ldots$, is—

(A) $28$ (B) $\mathbf{22}$ (C) $-38$ (D) $-48\frac{1}{2}$

Solution:
Given AP: $-3, -\frac{1}{2}, 2, \dots$
Here, First term ($a$) = $-3$
Common difference ($d$) = $-\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = \frac{-1+6}{2} = \frac{5}{2}$
Number of terms ($n$) = $11$
We know that, $a_n = a + (n - 1)d$
$\therefore a_{11} = -3 + (11 - 1)\left(\frac{5}{2}\right)$
$= -3 + 10 \times \frac{5}{2}$
$= -3 + 5 \times 5$
$= -3 + 25 = \mathbf{22}$
Hence, the correct choice is (B).

3. In the following APs, find the missing terms in the boxes—

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

(i) $2,\; \square,\; 26$

Here, $a = 2$ and $a_3 = 26$.
We know that $a_3 = a + 2d$
$26 = 2 + 2d \Rightarrow 24 = 2d \Rightarrow d = 12$
$\therefore$ Second term ($a_2$) = $a + d = 2 + 12 = \mathbf{14}$
Answer: $2,\; \mathbf{14},\; 26$

(ii) $\square,\; 13,\; \square,\; 3$

Here, $a_2 = 13$ and $a_4 = 3$.
$a + d = 13$ ----(1)
$a + 3d = 3$ ----(2)
Subtracting (1) from (2): $2d = -10 \Rightarrow d = -5$
Substitute $d = -5$ in (1): $a + (-5) = 13 \Rightarrow a = 18$
$\therefore a_3 = a + 2d = 18 + 2(-5) = \mathbf{8}$
Answer: $\mathbf{18},\; 13,\; \mathbf{8},\; 3$

(iii) $5,\; \square,\; \square,\; 9\frac{1}{2}$

Here, $a = 5$ and $a_4 = 9\frac{1}{2} = \frac{19}{2}$.
$a + 3d = \frac{19}{2} \Rightarrow 5 + 3d = \frac{19}{2}$
$3d = \frac{19}{2} - 5 = \frac{9}{2} \Rightarrow d = \frac{3}{2}$
$\therefore a_2 = 5 + \frac{3}{2} = \mathbf{6\frac{1}{2}}$ and $a_3 = 6\frac{1}{2} + \frac{3}{2} = \mathbf{8}$
Answer: $5,\; \mathbf{6\frac{1}{2}},\; \mathbf{8},\; 9\frac{1}{2}$

(iv) $-4,\; \square,\; \square,\; \square,\; \square,\; 6$

Here, $a = -4$ and $a_6 = 6$.
$a + 5d = 6 \Rightarrow -4 + 5d = 6 \Rightarrow 5d = 10 \Rightarrow d = 2$
Missing terms: $-4+2=\mathbf{-2}$, $-2+2=\mathbf{0}$, $0+2=\mathbf{2}$, $2+2=\mathbf{4}$
Answer: $-4,\; \mathbf{-2},\; \mathbf{0},\; \mathbf{2},\; \mathbf{4},\; 6$

(v) $\square,\; 38,\; \square,\; \square,\; \square,\; -22$

Here, $a_2 = 38$ and $a_6 = -22$.
$a + d = 38$ ----(1)
$a + 5d = -22$ ----(2)
Subtracting (1) from (2): $4d = -60 \Rightarrow d = -15$
Substitute $d = -15$ in (1): $a - 15 = 38 \Rightarrow a = 53$
Missing terms: $a_3 = 23$, $a_4 = 8$, $a_5 = -7$
Answer: $\mathbf{53},\; 38,\; \mathbf{23},\; \mathbf{8},\; \mathbf{-7},\; -22$

4. Which term of the AP: $3,\; 8,\; 13,\; 18,\; \ldots$ is $78$?

Solution:
Here, first term $a = 3$
Common difference $d = 8 - 3 = 5$
Let the $n$-th term be $a_n = 78$
We know that,
$a_n = a + (n - 1)d$
$\Rightarrow 78 = 3 + (n - 1)5$
$\Rightarrow 78 - 3 = (n - 1)5$
$\Rightarrow 75 = (n - 1)5$
$\Rightarrow n - 1 = \frac{75}{5}$
$\Rightarrow n - 1 = 15$
$\Rightarrow n = 15 + 1$
$\Rightarrow n = 16$
$\therefore$ The 16th term is 78.

5. Find the number of terms in each of the following APs:

(i) $7,\; 13,\; 19,\; \ldots,\; 205$

Solution:
Here, $a = 7, d = 6, a_n = 205$
$\Rightarrow a + (n - 1)d = a_n$
$\Rightarrow 7 + (n - 1)6 = 205$
$\Rightarrow (n - 1)6 = 205 - 7$
$\Rightarrow (n - 1)6 = 198$
$\Rightarrow n - 1 = \frac{198}{6}$
$\Rightarrow n - 1 = 33$
$\Rightarrow n = 34$

(ii) $18,\; 15\frac{1}{2},\; 13,\; \ldots,\; -47$

Solution:
Here, $a = 18, d = 15.5 - 18 = -2.5, a_n = -47$
$\Rightarrow 18 + (n - 1)(-2.5) = -47$
$\Rightarrow (n - 1)(-2.5) = -47 - 18$
$\Rightarrow (n - 1)(-2.5) = -65$
$\Rightarrow n - 1 = \frac{-65}{-2.5}$
$\Rightarrow n - 1 = 26$
$\Rightarrow n = 27$

6. Check whether $-150$ is a term of the AP: $11,\; 8,\; 5,\; 2,\; \ldots$

Solution:
Here, $a = 11, d = 8 - 11 = -3, a_n = -150$
$\Rightarrow a + (n - 1)d = -150$
$\Rightarrow 11 + (n - 1)(-3) = -150$
$\Rightarrow (n - 1)(-3) = -150 - 11$
$\Rightarrow (n - 1)(-3) = -161$
$\Rightarrow n - 1 = \frac{161}{3}$
$\Rightarrow n - 1 = 53.66$
Since the value of $n$ is not a positive integer, $-150$ cannot be a term of this AP.

7. Find the $31$st term of an AP whose $11$th term is $38$ and $16$th term is $73$.

Solution:
Let the first term be $a$ and common difference be $d$.
$a_{11} = a + 10d = 38$ ---(i)
$a_{16} = a + 15d = 73$ ---(ii)
Subtracting (i) from (ii), we get:
$\Rightarrow 5d = 73 - 38$
$\Rightarrow 5d = 35$
$\Rightarrow d = 7$
Substituting $d = 7$ in equation (i):
$\Rightarrow a + 10(7) = 38$
$\Rightarrow a + 70 = 38$
$\Rightarrow a = 38 - 70 = -32$
$\therefore a_{31} = a + 30d$
$\Rightarrow a_{31} = -32 + 30(7)$
$\Rightarrow a_{31} = -32 + 210$
$\Rightarrow a_{31} = 178$

8. An AP consists of $50$ terms of which the third term is $12$ and the last term is $106$. Find the $29$th term.

Solution:
Here, $n = 50, a_3 = 12, a_{50} = 106$
$\Rightarrow a + 2d = 12$ ---(i)
$\Rightarrow a + 49d = 106$ ---(ii)
Subtracting (i) from (ii), we get:
$\Rightarrow 47d = 94$
$\Rightarrow d = 2$
$a + 2(2) = 12 \Rightarrow a = 8$
$\therefore a_{29} = a + 28d$
$\Rightarrow a_{29} = 8 + 28(2)$
$\Rightarrow a_{29} = 8 + 56 = 64$

9. If the third and ninth terms of an AP are $4$ and $-8$ respectively, which term of this AP is zero?

Solution:
$a + 2d = 4$ ---(i)
$a + 8d = -8$ ---(ii)
Subtracting (i) from (ii), we get:
$\Rightarrow 6d = -12 \Rightarrow d = -2$
$a + 2(-2) = 4 \Rightarrow a = 8$
Let $a_n = 0$
$\Rightarrow 8 + (n - 1)(-2) = 0$
$\Rightarrow (n - 1)(-2) = -8$
$\Rightarrow n - 1 = 4$
$\Rightarrow n = 5$

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

Solution:
According to the question,
$a_{17} - a_{10} = 7$
$\Rightarrow (a + 16d) - (a + 9d) = 7$
$\Rightarrow 7d = 7$
$\Rightarrow d = 1$

11. Which term of the AP: $3,\; 15,\; 27,\; 39,\; \ldots$ will be $132$ more than its $54$th term?

Solution:
Here, $a = 3, d = 12$
Let the required term be $a_n$.
According to the question,
$a_n = a_{54} + 132$
$\Rightarrow a + (n - 1)d = a + 53d + 132$
$\Rightarrow 3 + (n - 1)(12) = 3 + 53 \times 12 + 132$
$\Rightarrow 3 + 12n - 12 = 3 + 636 + 132$
$\Rightarrow 12n - 9 = 771$
$\Rightarrow 12n = 771 + 9$
$\Rightarrow 12n = 780$
$\Rightarrow n = \frac{780}{12}$
$\Rightarrow n = 65$

12. Two APs have the same common difference. The difference between their $100$th terms is $100$. What is the difference between their $1000$th terms?

Solution:
Since the common difference $d$ is the same, the difference between any two corresponding terms at the same position remains constant.
Therefore, if the difference between the 100th terms is 100, the difference between the 1000th terms will also be 100.

13. How many three-digit numbers are divisible by $7$?

Solution:
The sequence is: $105, 112, 119, \dots, 994$
Here, $a = 105, d = 7, a_n = 994$
$\Rightarrow 994 = 105 + (n - 1)7$
$\Rightarrow 889 = (n - 1)7$
$\Rightarrow n - 1 = \frac{889}{7}$
$\Rightarrow n - 1 = 127$
$\Rightarrow n = 128$

14. How many multiples of $4$ lie between 10 and 250?

Solution:
The sequence is: $12, 16, 20, \dots, 248$
Here, $a = 12, d = 4, a_n = 248$
$\Rightarrow 248 = 12 + (n - 1)4$
$\Rightarrow 236 = (n - 1)4$
$\Rightarrow n - 1 = \frac{236}{4}$
$\Rightarrow n - 1 = 59$
$\Rightarrow n = 60$

15. For what value of n are the $n$-th terms of two APs: $63,\; 65,\; 67,\; \ldots$ and $3,\; 10,\; 17,\; \ldots$ equal?

Solution:
First AP: $a = 63, d = 2 \Rightarrow a_n = 63 + (n - 1)2$
Second AP: $a = 3, d = 7 \Rightarrow a_n = 3 + (n - 1)7$
According to the question,
$\Rightarrow 63 + (n - 1)2 = 3 + (n - 1)7$
$\Rightarrow 63 - 3 = (n - 1)7 - (n - 1)2$
$\Rightarrow 60 = (n - 1)5$
$\Rightarrow n - 1 = 12$
$\Rightarrow n = 13$

16. Determine the AP whose third term is $16$ and the $7$th term exceeds the $5$th term by $12$.

Solution:
According to the question, $a_7 - a_5 = 12$
$\Rightarrow (a + 6d) - (a + 4d) = 12$
$\Rightarrow 2d = 12 \Rightarrow d = 6$
Now, $a_3 = 16$
$\Rightarrow a + 2d = 16$
$\Rightarrow a + 2(6) = 16$
$\Rightarrow a + 12 = 16$
$\Rightarrow a = 4$
$\therefore$ The AP is: $4, 10, 16, 22, \dots$

17. Find the $20$th term from the last term of the AP: $3,\; 8,\; 13,\; \ldots,\; 253$.

Solution:
To find the term from the last, we write the AP in reverse order:
$253, 248, \dots, 13, 8, 3$
Here, $a = 253, d = -5$
$\Rightarrow a_{20} = a + 19d$
$\Rightarrow a_{20} = 253 + 19(-5)$
$\Rightarrow a_{20} = 253 - 95$
$\Rightarrow a_{20} = 158$

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Let the first term of the AP = $a$
Common difference = $d$

According to the first condition,

$a_4 + a_8 = 24$

$\Rightarrow a + 3d + a + 7d = 24$
$\Rightarrow 2a + 10d = 24$
$\Rightarrow a + 5d = 12$
$\Rightarrow a = 12 - 5d \dots \dots (i)$

According to the second condition,

$a_6 + a_{10} = 44$

$\Rightarrow a + 5d + a + 9d = 44$
$\Rightarrow 2a + 14d = 44$
$\Rightarrow a + 7d = 22 \dots \dots (ii)$

Substituting the value of $(a)$ from $(i)$ into $(ii)$,

$12 - 5d + 7d = 22$

$\Rightarrow 2d = 10$
$\Rightarrow d = \frac{10}{2}$
$\Rightarrow d = 5$

Substituting the value of $d$ in $(i)$,

$a = 12 - 5 \times 5$

$= 12 - 25$
$= -13$

$\therefore$ The first three terms of the AP are:
$a, a + d, a + 2d = -13, -8, -3$

19. 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 $Rs. 7000$?

Solution:
Here, $a = 5000, d = 200, a_n = 7000$
$\Rightarrow 7000 = 5000 + (n - 1)200$
$\Rightarrow 2000 = (n - 1)200$
$\Rightarrow n - 1 = 10$
$\Rightarrow n = 11$
In the 11th year, which is 1995 + 10 = 2005.

20. Ramkali saved $Rs. 5$ in the first week of a year and then increased her weekly savings by $Rs. 1.75$. If in the $n$-th week, her weekly savings become $Rs. 20.75$, find $n$.

Solution:
Here, $a = 5, d = 1.75, a_n = 20.75$
$\Rightarrow 20.75 = 5 + (n - 1)1.75$
$\Rightarrow 15.75 = (n - 1)1.75$
$\Rightarrow n - 1 = \frac{15.75}{1.75}$
$\Rightarrow n - 1 = 9$
$\Rightarrow n = 10$

21. Match Column (I) with Column (II).

Column (I)	Column (II)
P) The 5th term ($a_5$) of the AP $2,\;4,\;6,\;8,\ldots$ is	1) $-2$
Q) The common difference $(d)$ of the AP $-1.2,\;-3.2,\;-5.2,\;-7.2,\ldots$ is	2) $-4$
R) The first term $(a)$ of the AP $3,\;6,\;9,\;12,\ldots$ is	3) $10$
S) For the AP $0,\;-4,\;-8,\;-12,\ldots$, the value of $a_4 - a_3$ is	4) $3$

Choose the correct matching:

(A) P → 1, Q → 4, R → 3, S → 2

(B) P → 3, Q → 1, R → 4, S → 2

(C) P → 2, Q → 1, R → 4, S → 3

(D) P → 4, Q → 3, R → 2, S → 1

Solution:
P: Here $a=2, d=2$. So $a_5 = 2 + (5-1)2 = 10$. (Matches 3)
Q: Common difference $d = a_2 - a_1 = -3.2 - (-1.2) = -2$. (Matches 1)
R: The first term is clearly 3. (Matches 4)
S: $a_4 - a_3 = d = -4 - 0 = -4$. (Matches 2)
Correct Option: (B)

22. Which of the following statements are True or False?

P) $a_1,a_2,a_3,\ldots,a_n,\ldots$ will be an AP if $a_{n+1}-a_n$ is independent of $n$.

Q) If $d$ is the common difference of the AP $a_1, a_2, a_3, \ldots$, then $a_p-a_q=(p-q)d$.

R) In an AP with first term $a$ and common difference $d$, the $n$-th term is $a_n=a+(n-1)d$.

(A) P is false, Q and R are true
(B) P and Q are true, R is false
(C) P, Q, and R are all true
(D) P, Q, and R are all false

Solution: (C) All statements are correct mathematical properties of an Arithmetic Progression.

23. The steps to calculate the total number of terms in the AP $-1,\;3,\;7,\;11,\ldots,95$ are shuffled below:

(i) $-1+4n-4=95$

(ii) $n=25$

(iii) $a_n=95$

(iv) $-1+(n-1)\times4=95$

Choose the option that represents the correct logical sequence of the solution:

(A) (i)$\rightarrow$(iii)$\rightarrow$(ii)$\rightarrow$(iv)
(B) (ii)$\rightarrow$(i)$\rightarrow$(iii)$\rightarrow$(iv)
(C) (iii)$\rightarrow$(iv)$\rightarrow$(i)$\rightarrow$(ii)
(D) (iii)$\rightarrow$(iv)$\rightarrow$(ii)$\rightarrow$(i)

Solution: (C) First, identify the last term (iii), apply the general formula (iv), simplify the linear equation (i), and solve for $n$ (ii).

24. If the first three terms of an Arithmetic Progression are $b,\;c,\;2b$ respectively, determine the ratio $b:c$.

Solution:
Since the terms $b, c, 2b$ are in AP, the common difference must be the same.
$d = c - b$ $\dots (i)$
$d = 2b - c$ $\dots (ii)$

Equating (i) and (ii):
$c - b = 2b - c$
$\Rightarrow c + c = 2b + b$
$\Rightarrow 2c = 3b$
$\Rightarrow \frac{b}{c} = \frac{2}{3}$

Therefore, $b:c = 2:3$.

25. The digits of a three-digit positive number form an AP and their sum is $15$. The number obtained by reversing the digits is $594$ less than the original number. Find the number.

Solution:
Let the three digits be $(a-d), a, (a+d)$.
Sum of digits: $(a-d) + a + (a+d) = 15 \Rightarrow 3a = 15 \Rightarrow a = 5$.
Original number = $100(a-d) + 10a + (a+d)$
Reversed number = $100(a+d) + 10a + (a-d)$
According to the problem:
$(Original) - (Reversed) = 594$
$[100(5-d) + 50 + (5+d)] - [100(5+d) + 50 + (5-d)] = 594$
$\Rightarrow (555 - 99d) - (555 + 99d) = 594$
$\Rightarrow -198d = 594 \Rightarrow d = -3$.
The digits are: $5 - (-3) = 8$, $5$, and $5 + (-3) = 2$.
The number is 852.

Sudev Chandra Das (B.Sc. Mathematics)

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Important Points for HSLC Exam

Students should remember:

nth Term Formula
Common Difference
AP Identification
Pattern Questions
Formula Application

These topics are frequently asked in HSLC board examinations.

Common Difference Formula

$d=a_{n+1}-a_n$

Real-Life Uses of Arithmetic Progression

Arithmetic Progression is used in:

Banking calculations
Salary increment
Monthly savings
Business mathematics
Computer programming
Competitive examinations

Benefits of These Solutions

✅ Based on latest SEBA/SCERT Assam 2026–2027 syllabus
✅ Easy English explanations
✅ Step-by-step solutions
✅ HSLC exam focused preparation
✅ Important formulas included
✅ Student-friendly learning method

Final Words

These Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 Solutions help students understand AP concepts in a simple and easy way. Regular practice of all questions will improve problem-solving skills and help students score high marks in HSLC Examination.

Students should practice:

nth Term Formula
Common Difference
AP Pattern Questions
Formula-based Problems

daily for better understanding and strong mathematical foundation.

Class 10 Maths Chapter 5 – Arithmetic Progressions (Exercise 5.2) FAQs

Exercise 5.2 focuses on finding the nth term of an Arithmetic Progression (AP).

👉 Using the AP formula, students can easily calculate any term in the sequence.

An Arithmetic Progression is a sequence in which the difference between consecutive terms remains constant.

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📘 Here, a = first term and d = common difference.

The formula for finding the nth term of an Arithmetic Progression is:

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This formula helps in calculating any term quickly and accurately.

Yes, Exercise 5.2 is very important for HSLC exams. Questions worth 2–5 marks are frequently asked from this topic.

🎯 Learning the formula properly can help students score full marks easily.

The difference between two consecutive terms of an Arithmetic Progression is called the common difference (d).

Students often make mistakes while calculating the common difference or using (n−1) incorrectly in the formula.

🚀 Pro Tip: Identify the values of a, n, and d carefully before applying the formula.

Practice regularly, revise formulas daily, and solve previous year question papers for better speed and accuracy.

⭐ Consistent Maths practice improves confidence and exam performance.

Digital Pipal Academy provides step-by-step solutions, easy explanations, updated syllabus content, and exam-oriented preparation.

🎓 It helps students improve concepts, accuracy, and confidence in Mathematics.

You can find complete step-by-step solutions for Exercise 5.2 on Digital Pipal Academy.

Column (I)	Column (II)
P) The 5th term (\(a_5\)) of the AP \(2,\;4,\;6,\;8,\ldots\) is	1) \(-2\)
Q) The common difference \((d)\) of the AP \(-1.2,\;-3.2,\;-5.2,\;-7.2,\ldots\) is	2) \(-4\)
R) The first term \((a)\) of the AP \(3,\;6,\;9,\;12,\ldots\) is	3) \(10\)
S) For the AP \(0,\;-4,\;-8,\;-12,\ldots\), the value of \(a_4 - a_3\) is	4) \(3\)

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Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 Solutions | English Medium | SEBA/SCERT Assam 2026–2027