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Arithmetic Progressions: 3 marks Questions (CBSE Class 10)

34 different 3 marks questions on Arithmetic Progressions from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (77)2 marks (35)3 marks (34)4 marks (18)5 marks (38)

Q283 marksShort AnswerArithmetic Progressions CBSE 2026 · Basic 430/5/1

In an A.P., it is given that a = 2, d = 8 and $S_{n} = 90$ . Find the value of n.

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Answer: n = 5

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [4 + 8 (n - 1)] = 4 n^{2} - 2 n$
$4 n^{2} - 2 n = 90 \Rightarrow 2 n^{2} - n - 45 = 0$
$(2 n + 9) (n - 5) = 0$
n cannot be negative, so $n = 5$ .

Q28 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2026 · Basic 430/5/1

How many 4-digit numbers are divisible by 7 ?

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Answer: 1286

The smallest 4-digit multiple of 7 is 1001 and the largest is 9996.
1001, 1008, ..., 9996 is an A.P. with a = 1001, d = 7.
$9996 = 1001 + (n - 1) \times 7 \Rightarrow n - 1 = 1285 \Rightarrow n = 1286$

Q283 marksShort AnswerArithmetic Progressions CBSE 2026 · Basic 430/5/2

How many terms of the A.P. 3, 5, 7, 9, ... must be added to get the sum 80 ?

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Answer: 8 terms

a = 3, d = 2.
$S_{n} = \frac{n}{2} [6 + 2 (n - 1)] = n (n + 2) = 80$
$n^{2} + 2 n - 80 = 0 \Rightarrow (n + 10) (n - 8) = 0$
n cannot be negative, so n = 8.

Q28 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2026 · Basic 430/5/2

Find three consecutive terms in A.P. whose sum is 21 and their product is 231.

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Answer: 3, 7, 11 (or 11, 7, 3)

Let the terms be a − d, a, a + d.
Sum: 3a = 21, so a = 7.
Product: $(7 - d) \times 7 \times (7 + d) = 231 \Rightarrow 49 - d^{2} = 33 \Rightarrow d^{2} = 16$ , $d = \pm 4$
Terms: 3, 7, 11 (or 11, 7, 3).

Q273 marksShort AnswerArithmetic Progressions CBSE 2026 · Basic 430/5/3

The $4^{t h}$ and $1 0^{t h}$ term of an A.P. are 13 and 25 respectively. Find its $2 4^{t h}$ term.

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Answer: 53

a + 3d = 13 and a + 9d = 25.
Subtracting: 6d = 12, so d = 2 and a = 7.
$a_{24} = 7 + 23 \times 2 = 53$

Q27 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2026 · Basic 430/5/3

From 232 to 540, find the number of multiples of 3.

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Answer: 103

The first multiple of 3 after 232 is 234 and the last is 540.
234, 237, ..., 540 is an A.P. with a = 234, d = 3.
$540 = 234 + (n - 1) \times 3 \Rightarrow n - 1 = 102 \Rightarrow n = 103$

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

Q273 marksShort AnswerArithmetic Progressions CBSE 2025 · Basic 430/4/1

Find the sum of the A.P. $7, 10 \frac{1}{2}, 14, .....84$ .

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Answer:

\frac{2093}{2} = 1046 \frac{1}{2}

$a = 7$ , $d = 10 \frac{1}{2} - 7 = \frac{7}{2}$ , last term $l = 84$ .
$84 = 7 + (n - 1) \frac{7}{2}$ , so $n - 1 = 22$ and $n = 23$ .
$S_{23} = \frac{n}{2} (a + l) = \frac{23}{2} (7 + 84) = \frac{23 \times 91}{2} = \frac{2093}{2} = 1046 \frac{1}{2}$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q27 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2025 · Basic 430/4/1

If the sum of first $n$ terms of an A.P. is given by $S_{n} = \frac{n}{2} (2 n + 8)$ . Then, find its first term and common difference. Hence, find its $1 5^{t h}$ term.

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Answer: First term 5, common difference 2,

a_{15} = 33

$S_{n} = \frac{n}{2} (2 n + 8) = n^{2} + 4 n$ .
$a_{1} = S_{1} = 1 + 4 = 5$ .
$S_{2} = 4 + 8 = 12$ , so $a_{2} = S_{2} - S_{1} = 7$ and $d = 7 - 5 = 2$ .
$a_{15} = a + 14 d = 5 + 28 = 33$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q313 marksShort AnswerArithmetic Progressions CBSE 2025 · Basic 430/6/1

Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.

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Answer: A.P.: 4, 10, 16, 22, ...;

S_{29} = 2552

$a_{7} - a_{5} = 2 d = 12$ , so $d = 6$ .
$a_{3} = a + 2 d = 16$ , so $a = 4$ .
A.P.: 4, 10, 16, 22, ...
$S_{29} = \frac{29}{2} [2 \times 4 + 28 \times 6] = \frac{29}{2} \times 176 = 2552$ .

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q31 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2025 · Basic 430/6/1

Find the sum of first 20 terms of an A.P. whose $n^{t h}$ term is given by $a_{n} = 5 + 2 n$ . Can 52 be a term of this A.P. ?

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Answer:

S_{20} = 520

; No, 52 is not a term.

$a_{1} = 7$ , $a_{20} = 5 + 40 = 45$ .
$S_{20} = \frac{20}{2} (7 + 45) = 520$ .
If $5 + 2 n = 52$ , then $n = \frac{47}{2}$ , which is not a natural number.
So 52 is not a term of the A.P.

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q313 marksShort AnswerArithmetic Progressions CBSE 2025 · Standard 30/2/1

A sum of ₹ 2,000 is invested at 7% per annum simple interest. Calculate the interests at the end of $1^{s t}$ , $2^{n d}$ and $3^{r d}$ year. Do these interests form an AP ? If so, find the interest at the end of the $2 7^{t h}$ year.

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Answer: ₹ 140, ₹ 280, ₹ 420; yes, they form an AP; interest at the end of the 27th year = ₹ 3,780

Interest for one year $= \frac{2000 \times 7 \times 1}{100} = ₹ 140$
Interest at end of years 1, 2, 3: ₹ 140, ₹ 280, ₹ 420
Differences are equal (140), so they form an AP with $a = 140$ , $d = 140$
$a_{27} = 140 + 26 \times 140 = 3780$ , i.e. ₹ 3,780

Also asked in: 2025 Standard 30/2/2

Q273 marksShort AnswerArithmetic Progressions CBSE 2025 · Standard 30/2/3

Find the sum of all 3-digit natural numbers which are divisible by 11.

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Answer: 44550

3-digit multiples of 11: 110, 121, ..., 990, an AP with $a = 110$ , $d = 11$
$990 = 110 + (n - 1) \times 11$ , so $n - 1 = 80$ , $n = 81$
$S_{81} = \frac{81}{2} (110 + 990) = 81 \times 550 = 44550$

Q263 marksShort AnswerArithmetic Progressions CBSE 2024 · Standard 30/2/1

If the sum of first m terms of an A.P. is same as sum of its first n terms $(m \neq = n)$ , then show that the sum of its first $(m + n)$ terms is zero.

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Answer: Proved.

$S_{m} = S_{n}$ gives $\frac{m}{2} [2 a + (m - 1) d] = \frac{n}{2} [2 a + (n - 1) d]$ .
$2 a (m - n) + d [(m^{2} - n^{2}) - (m - n)] = 0$
$(m - n) [2 a + (m + n - 1) d] = 0$
Since $m \neq = n$ , $2 a + (m + n - 1) d = 0$ .
$S_{m + n} = \frac{m + n}{2} [2 a + (m + n - 1) d] = \frac{m + n}{2} \times 0 = 0$

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

Q26 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2024 · Standard 30/2/1

In an A.P., the sum of three consecutive terms is 24 and the sum of their squares is 194. Find the numbers.

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Answer: 7, 8, 9 (or 9, 8, 7)

Let the terms be $a - d$ , $a$ , $a + d$ .
Sum: $3 a = 24$ , so $a = 8$ .
Squares: $(8 - d)^{2} + 64 + (8 + d)^{2} = 194$
$192 + 2 d^{2} = 194$ , so $d^{2} = 1$ and $d = \pm 1$ .
The numbers are 7, 8, 9 (or 9, 8, 7).

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

Q263 marksShort AnswerArithmetic Progressions CBSE 2024 · Standard 30/3/1

If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first 20 terms.

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Answer:

S_{20} = 400

$S_{7} = \frac{7}{2} (2 a + 6 d) = 49 \Rightarrow a + 3 d = 7$ .
$S_{17} = \frac{17}{2} (2 a + 16 d) = 289 \Rightarrow a + 8 d = 17$ .
Subtracting: $5 d = 10$ , so $d = 2$ and $a = 1$ .
$S_{20} = \frac{20}{2} (2 \times 1 + 19 \times 2) = 10 \times 40 = 400$ .

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

Q26 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2024 · Standard 30/3/1

The ratio of the $1 0^{t h}$ term to its $3 0^{t h}$ term of an A.P. is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference of A.P.

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Answer: First term

a = 2

, common difference

d = 2

$\frac{a + 9 d}{a + 29 d} = \frac{1}{3} \Rightarrow 3 a + 27 d = a + 29 d \Rightarrow a = d$ .
$S_{6} = \frac{6}{2} (2 a + 5 d) = 42 \Rightarrow 2 a + 5 d = 14$ .
With $a = d$ : $7 a = 14$ , so $a = 2$ and $d = 2$ .

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

Q313 marksShort AnswerArithmetic Progressions CBSE 2024 · Standard 30/4/2

A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was ₹ 15,000 after 4 years of service and ₹ 18,000 after 10 years of service, what was his starting salary and what was the annual increment ?

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Answer: Starting salary ₹ 13,000; annual increment ₹ 500

Let the starting salary be ₹ $a$ and the annual increment ₹ $d$ ; after $n$ years the salary is $a + n d$ .
$a + 4 d = 15000$ and $a + 10 d = 18000$ .
Subtracting: $6 d = 3000$ , so $d = 500$ .
$a = 15000 - 2000 = 13000$ .
Starting salary ₹ 13,000 and annual increment ₹ 500.

Q273 marksShort AnswerArithmetic Progressions CBSE 2024 · Standard 30/5/1

If the sum of the first 14 terms of an A.P. is 1050 and the first term is 10, then find the $2 0^{t h}$ term and the $n^{t h}$ term.

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Answer:

a_{20} = 200

a_{n} = 10 n

$S_{14} = \frac{14}{2} [2 \times 10 + 13 d] = 1050 \Rightarrow 20 + 13 d = 150 \Rightarrow d = 10$ .
$a_{20} = 10 + 19 \times 10 = 200$ .
$a_{n} = 10 + (n - 1) 10 = 10 n$ .

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

Q263 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/4/1

How many terms are there in an A.P. whose first and fifth terms are $- 14$ and 2, respectively and the last term is 62.

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Answer: 20 terms

$a = - 14$ , $a_{5} = a + 4 d = 2 \Rightarrow 4 d = 16 \Rightarrow d = 4$
Last term: $a + (n - 1) d = 62$
$- 14 + 4 (n - 1) = 62 \Rightarrow n - 1 = 19 \Rightarrow n = 20$

Q26 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/4/1

Which term of the A.P. : 65, 61, 57, 53, .................. is the first negative term ?

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Answer: 18th term

$a = 65$ , $d = - 4$
$a_{n} = 65 + (n - 1) (- 4) = 69 - 4 n$
$a_{n} < 0 \Rightarrow 69 - 4 n < 0 \Rightarrow n > 17.25$
So the first negative term is the 18th term ( $a_{18} = - 3$ ).

Also asked in: 2023 Standard 30/4/3

Q293 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/4/2

The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its $2 0^{t h}$ term.

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Answer: 110

$S_{15} = \frac{15}{2} [2 (15) + 14 d] = 750$
$30 + 14 d = 100 \Rightarrow d = 5$
$a_{20} = a + 19 d = 15 + 95 = 110$

Q29 (OR) (OR)3 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/4/2

Rohan repays his total loan of ₹ 1,18,000 by paying every month starting with the first instalment of ₹ 1,000. If he increases the instalment by ₹ 100 every month, what amount will be paid by him in the $3 0^{t h}$ instalment ? What amount of loan has he paid after $3 0^{t h}$ instalment ?

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Answer: 30th instalment = ₹3,900; amount paid after 30 instalments = ₹73,500

Instalments form an A.P. with $a = 1000$ , $d = 100$ .
$a_{30} = 1000 + 29 \times 100 = 3900$ , so the 30th instalment is ₹3,900.
$S_{30} = \frac{30}{2} (1000 + 3900) = 15 \times 4900 = 73500$
Amount paid after the 30th instalment = ₹73,500 (₹44,500 still remains).

Q293 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/4/3

How many terms are there in A.P. whose first and fifth term are $- 14$ and 2, respectively and the last term is 62.

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Answer: 20 terms

$a = - 14$ , $a_{5} = a + 4 d = 2 \Rightarrow 4 d = 16 \Rightarrow d = 4$
Last term: $a + (n - 1) d = 62$
$- 14 + 4 (n - 1) = 62 \Rightarrow n - 1 = 19 \Rightarrow n = 20$

Q273 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/5/1

If $p^{t h}$ term of an A.P. is q and $q^{t h}$ term is p, then prove that its $n^{t h}$ term is $(p + q - n)$ .

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Answer: Proved.

$a + (p - 1) d = q$ ... (1)
$a + (q - 1) d = p$ ... (2)
(1) – (2): $(p - q) d = q - p$ , so d = $- 1$ (p $\neq =$ q).
From (1): $a = q + (p - 1) = p + q - 1$ .
$a_{n} = a + (n - 1) d = p + q - 1 - (n - 1) = p + q - n$ .
Hence proved.

Q303 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/5/2

In an A.P., the sum of the first n terms is given by $S_{n} = 6 n - n^{2}$ . Find its $3 0^{t h}$ term.

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Answer:

a_{30} = - 53

$a_{30} = S_{30} - S_{29}$ .
$S_{30} = 180 - 900 = - 720$ ; $S_{29} = 174 - 841 = - 667$ .
$a_{30} = - 720 - (- 667) = - 53$ .

Q313 marksShort AnswerArithmetic Progressions CBSE 2023 · Standard 30/5/3

Find the common difference of an A.P. whose first term is 8, the last term is 65 and the sum of all its terms is 730.

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Answer: d = 3

$S_{n} = \frac{n}{2} (a + l)$ : $730 = \frac{n}{2} (8 + 65)$ , so $n = \frac{1460}{73} = 20$ .
$l = a + (n - 1) d$ : $65 = 8 + 19 d$ , so $d = \frac{57}{19} = 3$ .

Q93 marksShort AnswerArithmetic Progressions CBSE 2022 · Basic 430/4/1

Find the sum of first 16 terms of the A.P. whose n $^{th}$ term is given by $a_{n} = 5 n - 3$ .

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Answer:

S_{16} = 632

$a_{1} = 5 (1) - 3 = 2$ , $a_{16} = 5 (16) - 3 = 77$ .
$S_{16} = \frac{16}{2} (a_{1} + a_{16}) = 8 (2 + 77) = 8 \times 79 = 632$ .

Q83 marksShort AnswerArithmetic Progressions CBSE 2022 · Basic 430/4/2

In an A.P., the first term is 12 and the common difference is 6. If the last term of the A.P. is 252, then find its middle term.

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Answer: Middle term = 132

$a = 12$ , $d = 6$ , $a_{n} = 252$ .
$12 + (n - 1) 6 = 252 \Rightarrow n - 1 = 40 \Rightarrow n = 41$ .
Number of terms is odd, so the middle term is the $\frac{41 + 1}{2} = 21$ st term.
$a_{21} = 12 + 20 \times 6 = 132$ .

Q93 marksShort AnswerArithmetic Progressions CBSE 2022 · Basic 430/4/3

The sum of first n terms of an AP is given by $S_{n} = 3 n^{2} + 2 n$ . Find the AP.

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Answer: 5, 11, 17, 23, ...

$a_{1} = S_{1} = 3 + 2 = 5$ .
$S_{2} = 12 + 4 = 16$ , so $a_{2} = S_{2} - S_{1} = 11$ .
$d = 11 - 5 = 6$ .
The AP is 5, 11, 17, 23, ...

Q103 marksShort AnswerArithmetic Progressions CBSE 2022 · Standard 30/4/1

If the last term of an A.P. of 30 terms is 119 and the $8^{t h}$ term from the end (towards the first term) is 91, then find the common difference of the A.P. Hence, find the sum of all the terms of the A.P.

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Answer: