Arithmetic Progressions: CBSE Class 10 Previous Year Questions

202 different questions from Arithmetic Progressions (NCERT Chapter 5) asked in CBSE Class 10 Maths board exams 2022–2026. 2 of them came up in more than one year. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions77 questions 2 marks questions35 questions 3 marks questions34 questions 4 marks questions18 questions 5 marks questions38 questions

Most asked Arithmetic Progressions questions

Q325 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/6/2

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

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Answer: Number of terms = 16; common difference =

\frac{8}{3}

$S_{n} = \frac{n}{2} (a + l)$ : $400 = \frac{n}{2} (5 + 45) = 25 n$ , so $n = 16$ .
$l = a + (n - 1) d$ : $45 = 5 + 15 d$ , so $d = \frac{40}{15} = \frac{8}{3}$ .

Also asked in: 2024 Standard 30/5/1, 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q121 markMCQArithmetic Progressions CBSE 2024 · Standard 30/4/1

Three numbers in A.P. have the sum 30. What is its middle term ?

(A)4
(B)10
(C)16
(D)8

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Answer: (B) 10

Let the numbers be $a - d$ , $a$ , $a + d$ .
Sum = $3 a = 30$ , so $a = 10$ .
Middle term = 10.

Also asked in: 2025 Standard 30/1/2, 2024 Standard 30/4/3

Q141 markMCQArithmetic Progressions CBSE 2026 · Basic 430/4/1

$n^{t h}$ term of the A.P. $- \frac{3}{2}, \frac{3}{2}, \frac{9}{2}$ , ... is :

(A) $\frac{3 n}{2} - 3$
(B) $3 n - \frac{9}{2}$
(C) $\frac{3 n - 9}{2}$
(D) $3 n + \frac{3}{2}$

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Answer: (B)

3 n - \frac{9}{2}

$a = - \frac{3}{2}$ , $d = \frac{3}{2} - (- \frac{3}{2}) = 3$ .
$a_{n} = - \frac{3}{2} + (n - 1) 3 = 3 n - \frac{9}{2}$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q201 markAssertion–ReasonArithmetic Progressions CBSE 2026 · Standard 30/2/1

Assertion (A) : The mean of first 'n' natural numbers is $\frac{n - 1}{2}$ .
Reason (R): The sum of first 'n' natural numbers is $\frac{n ( n + 1 )}{2}$ .

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

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Answer: (D) Assertion (A) is false, but Reason (R) is true.

Sum of first n natural numbers $= \frac{n ( n + 1 )}{2}$ , so R is true.
Mean $= \frac{n ( n + 1 )}{2} \div n = \frac{n + 1}{2}$ , not $\frac{n - 1}{2}$ , so A is false.

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q111 markMCQArithmetic Progressions CBSE 2026 · Standard 30/3/1

The $n^{t h}$ term of the A.P. $\frac{- 1}{3}, \frac{2}{3}, \frac{5}{3}, \frac{8}{3}, ...$ is :

(A) $3 n - 4$
(B) $n - \frac{4}{3}$
(C) $\frac{n - 2}{3}$
(D) $\frac{n - 4}{3}$

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Answer: (B)

n - \frac{4}{3}

$a = - \frac{1}{3}$ , $d = \frac{2}{3} - (- \frac{1}{3}) = 1$
$a_{n} = - \frac{1}{3} + (n - 1) \times 1 = n - \frac{4}{3}$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q273 marksShort AnswerArithmetic Progressions CBSE 2026 · Standard 30/5/1

In an A.P., $1 5^{t h}$ term exceeds the $8^{t h}$ term by 21. If sum of first 10 terms is 55, then form the A.P.

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Answer:

- 8, - 5, - 2, 1, \dots

$a_{15} - a_{8} = 7 d = 21$ , so $d = 3$ .
$S_{10} = \frac{10}{2} [2 a + 9 d] = 5 (2 a + 27) = 55$
$2 a + 27 = 11$ , so $a = - 8$ .
A.P.: $- 8, - 5, - 2, 1, \dots$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q27 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2026 · Standard 30/5/1

The sum of first n terms of an A.P. is $2 n^{2} + 13 n$ . Find its $n^{t h}$ term and hence $1 0^{t h}$ term.

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Answer:

a_{n} = 4 n + 11

;

a_{10} = 51

$S_{n} = 2 n^{2} + 13 n$
$S_{n - 1} = 2 (n - 1)^{2} + 13 (n - 1) = 2 n^{2} + 9 n - 11$
$a_{n} = S_{n} - S_{n - 1} = 4 n + 11$
$a_{10} = 4 (10) + 11 = 51$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q364 marksCase StudyArithmetic Progressions CBSE 2026 · Standard 30/1/1

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato. The other potatoes are arranged 3 m apart in a straight line, with a total of 10 potatoes, as shown in the figure :
A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.
Based on the above information, answer the following questions :
(i) What is the distance covered to pick up the first potato and drop it in bucket ? (1)
(ii) What is the distance covered to pick up the second potato and drop it in bucket ? (1)
(iii) (a) What is the total distance the competitor has to run ? (2)
OR
(iii) (b) If average speed of competitor is 5 m/s, then find the average time taken by competitor to put all the potatoes in the bucket. (2)

Diagram for CBSE 2026 Class 10 Maths question 36

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Answer: (i) 10 m (ii) 16 m (iii) (a) 370 m OR (b) 74 s

The potatoes are 5 m, 8 m, 11 m, ..., from the bucket; the 10th is $5 + 9 \times 3 = 32$ m away.
(i) Distance = $2 \times 5 = 10$ m.
(ii) Distance = $2 \times 8 = 16$ m.
The round-trip distances 10, 16, 22, ... form an AP with $a = 10$ , $d = 6$ , $n = 10$ .
(iii) (a) Total $= \frac{10}{2} [2 \times 10 + 9 \times 6] = 5 \times 74 = 370$ m.
(iii) (b) Time $= \frac{370}{5} = 74$ s.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q364 marksCase StudyArithmetic Progressions CBSE 2026 · Standard 30/2/1

Your elder brother wants to buy a car and plans to take a loan from a bank for his car. He repays his total loan of ₹ 1,18,000 by paying every month, starting with the first instalment of ₹ 1,000 and he increases the instalment by ₹ 100 every month.
Based on the information given above, answer the following questions :
(i) Find the amount paid by him in the $3 0^{t h}$ instalment. (1)
(ii) If the total number of instalments is 40, what is the amount paid in the last instalment ? (1)
(iii) (a) What amount does he still have to pay after the $3 0^{t h}$ instalment ? (2)
OR
(iii) (b) Find the ratio of the tenth instalment to the last instalment. (2)

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Answer: (i) ₹ 3,900 (ii) ₹ 4,900 (iii) (a) ₹ 44,500 OR (b) 19 : 49

Instalments form an A.P. with a = 1000, d = 100.
(i) $a_{30} = 1000 + 29 \times 100 = 3900$ , i.e. ₹ 3,900
(ii) $a_{40} = 1000 + 39 \times 100 = 4900$ , i.e. ₹ 4,900
(iii)(a) $S_{30} = \frac{30}{2} [2 (1000) + 29 (100)] = 15 \times 4900 = 73500$
Amount still to pay = 118000 - 73500 = ₹ 44,500
(iii)(b) $a_{10} = 1000 + 9 \times 100 = 1900$ ; last instalment = 4900
Ratio = 1900 : 4900 = 19 : 49

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q374 marksCase StudyArithmetic Progressions CBSE 2026 · Standard 30/4/1

‘Kolam’ is a decorative art which is made with rice flour in South Indian States. It is drawn on grid pattern of dots. One such art work is shown below.
Observe the given figure carefully. There are 4 dots in first square, 8 dots in second square, 12 dots in third square and so on.
Based on the above, answer the following questions :
(i) Show that number of dots given above form an A.P. Write the first term and common difference. (1)
(ii) Write $n^{t h}$ term of the A.P. formed. (1)
(iii) (a) The pattern is expanded on a large ground. If total 220 dots are used, then find the number of squares formed. (2)
OR (b) Is it possible to complete n number of squares using 100 dots ? If yes, then find the value of n.

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Answer: (i) 4, 8, 12, ... is an A.P. with a = 4, d = 4 (ii)

a_{n} = 4 n

(iii) (a) 10 squares OR (b) No

(i) 8 $-$ 4 = 12 $-$ 8 = 4, a constant difference, so 4, 8, 12, ... is an A.P. with a = 4, d = 4.
(ii) $a_{n} = 4 + (n - 1) 4 = 4 n$
(iii)(a) $S_{n} = \frac{n}{2} [8 + 4 (n - 1)] = 2 n (n + 1) = 220$
$n^{2} + n - 110 = 0 \Rightarrow (n + 11) (n - 10) = 0 \Rightarrow n = 10$ squares
(iii)(b) $2 n (n + 1) = 100 \Rightarrow n^{2} + n - 50 = 0$
$D = 1 + 200 = 201$ is not a perfect square, so n is not a natural number. Not possible.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

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