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Arithmetic Progressions: 2 marks Questions (CBSE Class 10)

35 different 2 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)

Q222 marksVery Short AnswerArithmetic Progressions CBSE 2026 · Standard 30/3/1

In an A.P., the first term is 32 and the last term is – 10. If the common difference is – 2, then find the number of terms and their sum.

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Answer: 22 terms; sum = 242

$a = 32$ , $l = - 10$ , $d = - 2$
$l = a + (n - 1) d$ : $- 10 = 32 + (n - 1) (- 2)$ , so $n - 1 = 21$ , $n = 22$
$S = \frac{n}{2} (a + l) = \frac{22}{2} (32 - 10) = 11 \times 22 = 242$

Also asked in: 2026 Standard 30/3/2

Q22 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2026 · Standard 30/3/1

Find the sum of the first 28 terms of an A.P. whose $n^{t h}$ term is given by $a_{n} = 3 n - 2$ .

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Answer: 1162

$a_{1} = 3 (1) - 2 = 1$ , $a_{28} = 3 (28) - 2 = 82$
$S_{28} = \frac{28}{2} (1 + 82) = 14 \times 83 = 1162$

Also asked in: 2026 Standard 30/3/2

Q232 marksVery Short AnswerArithmetic Progressions CBSE 2026 · Standard 30/3/3

In an A.P., the first term is 4 and the last term is 31. If sum of all the terms is 175, find the number of terms and the common difference.

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Answer:

n = 10

d = 3

$S = \frac{n}{2} (a + l)$ : $175 = \frac{n}{2} (4 + 31)$ , so $n = 10$
$l = a + (n - 1) d$ : $31 = 4 + 9 d$ , so $d = 3$

Q23 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2026 · Standard 30/3/3

How many terms of the A.P. 21, 18, 15, .... must be added to get the sum zero ?

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Answer: 15

$a = 21$ , $d = - 3$
$S_{n} = \frac{n}{2} [2 \times 21 + (n - 1) (- 3)] = 0$
$42 - 3 n + 3 = 0$ (as $n \neq = 0$ ), so $n = 15$

Q252 marksVery Short AnswerArithmetic Progressions CBSE 2023 · Basic 430/5/1

Find the sum of the first 15 terms of the A.P. : $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, ......$ .

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Answer:

\frac{11}{4}

$a = \frac{1}{15}$ , $d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$ .
$S_{15} = \frac{15}{2} [2 \times \frac{1}{15} + 14 \times \frac{1}{60}] = \frac{15}{2} [\frac{8}{60} + \frac{14}{60}]$ .
$= \frac{15}{2} \times \frac{22}{60} = \frac{11}{4}$ .

Q212 marksVery Short AnswerArithmetic Progressions CBSE 2023 · Basic 430/5/2

Find the sum of the first 20 terms of the A.P. : $\frac{2}{3}, 0, - \frac{2}{3}, - \frac{4}{3}, ......$ .

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Answer:

- \frac{340}{3}

$a = \frac{2}{3}$ , $d = 0 - \frac{2}{3} = - \frac{2}{3}$ .
$S_{20} = \frac{20}{2} [2 \times \frac{2}{3} + 19 \times (- \frac{2}{3})] = 10 [\frac{4}{3} - \frac{38}{3}]$ .
$= 10 \times (- \frac{34}{3}) = - \frac{340}{3}$ .

Q222 marksVery Short AnswerArithmetic Progressions CBSE 2023 · Basic 430/5/3

Find the sum of the first 20 terms of the A.P. : $- \frac{29}{3}, - 9, - \frac{25}{3}, - \frac{23}{3}, ......$ .

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Answer:

- \frac{200}{3}

$a = - \frac{29}{3}$ , $d = - 9 + \frac{29}{3} = \frac{2}{3}$ .
$S_{20} = \frac{20}{2} [2 \times (- \frac{29}{3}) + 19 \times \frac{2}{3}] = 10 [- \frac{58}{3} + \frac{38}{3}]$ .
$= 10 \times (- \frac{20}{3}) = - \frac{200}{3}$ .

Q12 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/1/1

In an AP, if $a = 50$ , $d = - 4$ and $S_{n} = 0$ , then find the value of $n$ .

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Answer:

n = 26

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 0$
$\frac{n}{2} [100 + (n - 1) (- 4)] = 0$
Since $n \neq = 0$ , $100 - 4 n + 4 = 0$
$4 n = 104$ , so $n = 26$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q1 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/1/1

Find the sum of the first twelve 2-digit multiples of 7, using an AP.

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Answer: 630

The 2-digit multiples of 7 are 14, 21, 28, ... (an AP with $a = 14$ , $d = 7$ )
For $n = 12$ : $S_{12} = \frac{12}{2} [2 (14) + 11 (7)]$
$= 6 (28 + 77) = 6 \times 105 = 630$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q22 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/2/1

Which term of the A.P. 3, 8, 13, 18, … is 78 ?

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Answer: 16th term

$a = 3$ , $d = 8 - 3 = 5$ .
$a_{n} = a + (n - 1) d = 78$
$3 + 5 (n - 1) = 78$
$5 (n - 1) = 75 \Rightarrow n - 1 = 15 \Rightarrow n = 16$ .
So 78 is the 16th term.

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q2 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/2/1

Find the common difference of an A.P. whose $n^{t h}$ term is given by $a_{n} = 6 n - 5$ .

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Answer: Common difference = 6

$a_{1} = 6 (1) - 5 = 1$ , $a_{2} = 6 (2) - 5 = 7$ .
$d = a_{2} - a_{1} = 7 - 1 = 6$ .
(In general $a_{n + 1} - a_{n} = 6$ .)

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q62 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/2/1

Find the sum of the first fifteen multiples of 8.

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Answer: 960

Multiples: 8, 16, 24, …, 120 with $a = 8$ , $d = 8$ , $n = 15$ .
$S_{15} = \frac{15}{2} [2 (8) + 14 (8)] = \frac{15}{2} \times 128 = 960$ .

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q22 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/3/1

For an AP with common difference 6, the sum of first ten terms is same as four times the sum of first five terms. Determine the first term of the AP.

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Answer: First term

= 3

Let the first term be $a$ ; $d = 6$ .
$S_{10} = \frac{10}{2} [2 a + 9 (6)] = 10 a + 270$
$S_{5} = \frac{5}{2} [2 a + 4 (6)] = 5 a + 60$ , so $4 S_{5} = 20 a + 240$
$10 a + 270 = 20 a + 240 \Rightarrow 10 a = 30 \Rightarrow a = 3$

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q52 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/3/1

Which term of AP : $\frac{11}{4}$ , $\frac{7}{2}$ , $\frac{17}{4}$ , ... is $\frac{53}{4}$ ?

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Answer: 15th term

$a = \frac{11}{4}$ , $d = \frac{7}{2} - \frac{11}{4} = \frac{3}{4}$
$a_{n} = a + (n - 1) d = \frac{53}{4}$
$\frac{11}{4} + (n - 1) \frac{3}{4} = \frac{53}{4} \Rightarrow (n - 1) \frac{3}{4} = \frac{42}{4}$
$n - 1 = 14 \Rightarrow n = 15$

Q52 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/3/2

Find the $1 5^{t h}$ term of an AP whose first term is 17 and fourth term is 44.

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Answer:

a_{15} = 143

$a = 17$ , $a_{4} = a + 3 d = 44 \Rightarrow 3 d = 27 \Rightarrow d = 9$
$a_{15} = a + 14 d = 17 + 14 \times 9 = 17 + 126 = 143$

Q22 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/3/3

Find the $8^{t h}$ term of an AP whose first term is $- 24$ and $1 1^{t h}$ term is 21.

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Answer:

a_{8} = 7.5

$a = - 24$ , $a_{11} = a + 10 d = 21 \Rightarrow 10 d = 45 \Rightarrow d = 4.5$
$a_{8} = a + 7 d = - 24 + 7 (4.5) = - 24 + 31.5 = 7.5$

Q22 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/4/1

Determine the 36 $^{th}$ term of the A.P. whose first two terms are $- 3$ and 4 respectively.

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Answer:

a_{36} = 242

$a = - 3$ , $d = 4 - (- 3) = 7$ .
$a_{36} = a + 35 d = - 3 + 35 \times 7 = - 3 + 245 = 242$ .

Also asked in: 2022 Basic 430/4/2

Q2 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/4/1

Write the next two terms of the A.P. :
$27, 48, 75, .......$

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Answer:

108

and

147

(i.e.

63

and

73

)

$27 = 33$ , $48 = 43$ , $75 = 53$ .
So $d = 3$ .
Next two terms: $63 = 108$ and $73 = 147$ .

Also asked in: 2022 Basic 430/4/2

Q12 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/4/3

Find the 15 $^{th}$ term from the end (towards first term) of the A.P. 3, 8, 13, ........, 253.

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Answer: 183

$d = 8 - 3 = 5$ , last term $l = 253$ .
Reading from the end, the A.P. has first term 253 and common difference $- 5$ .
15th term from the end $= 253 + 14 (- 5) = 253 - 70 = 183$ .

Q1 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Basic 430/4/3

Write the next two terms of the A.P. : $5, 20, 45, ........$

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Answer:

80

and

125

(i.e.

45

and

55

)

$5$ , $20 = 25$ , $45 = 35$ .
So $d = 5$ .
Next two terms: $45 = 80$ and $55 = 125$ .

Q12 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/1/1

Find the sum of first 30 terms of AP : $- 30, - 24, - 18$ , ..... .

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Answer: 1710

Here $a = - 30$ , $d = - 24 - (- 30) = 6$ , $n = 30$ .
$S_{30} = \frac{30}{2} [2 (- 30) + (30 - 1) (6)]$
$= 15 [- 60 + 174] = 15 \times 114 = 1710$

Also asked in: 2022 Standard 30/1/2

Q1 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/1/1

In an AP if $S_{n} = n (4 n + 1)$ , then find the AP.

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Answer: 5, 13, 21, 29, ...

$S_{1} = 1 (4 + 1) = 5$ , so $a_{1} = 5$ .
$S_{2} = 2 (8 + 1) = 18$ , so $a_{2} = S_{2} - S_{1} = 13$ .
$d = 13 - 5 = 8$ .
The AP is 5, 13, 21, 29, ...

Also asked in: 2022 Standard 30/1/2

Q32 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/1/3

Find the sum of the first twelve 2-digit numbers which are multiples of 6.

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Answer: 540

The 2-digit multiples of 6 are 12, 18, 24, ...; $a = 12$ , $d = 6$ .
$a_{12} = 12 + 11 \times 6 = 78$ .
$S_{12} = \frac{12}{2} (12 + 78) = 6 \times 90 = 540$

Q3 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/1/3

In an AP, if $a_{2} = 26$ and $a_{15} = - 26$ , then write the AP.

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Answer: 30, 26, 22, 18, ...

$a + d = 26$ and $a + 14 d = - 26$ .
Subtracting: $13 d = - 52 \Rightarrow d = - 4$ .
$a = 26 - (- 4) = 30$ .
The AP is 30, 26, 22, 18, ...

Q22 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/2/1

Which term of the A.P. $- \frac{11}{2}, - 3, - \frac{1}{2},$ .... is $\frac{49}{2}$ ?

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Answer: 13th term

$a = - \frac{11}{2}$ , $d = - 3 - (- \frac{11}{2}) = \frac{5}{2}$ .
$a_{n} = a + (n - 1) d = \frac{49}{2}$
$- \frac{11}{2} + (n - 1) \frac{5}{2} = \frac{49}{2}$
$(n - 1) \frac{5}{2} = 30 \Rightarrow n - 1 = 12 \Rightarrow n = 13$ .
So $\frac{49}{2}$ is the 13th term.

Also asked in: 2022 Standard 30/2/2

Q2 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/2/1

Find a and b so that the numbers a, 7, b, 23 are in A.P.

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Answer:

a = - 1

b = 15

In an A.P. each middle term is the average of its neighbours.
7 is between a and b, and b is between 7 and 23.
$b = \frac{7 + 23}{2} = 15$ .
Common difference $d = 15 - 7 = 8$ , so $a = 7 - 8 = - 1$ .
Check: $- 1, 7, 15, 23$ has common difference 8.

Also asked in: 2022 Standard 30/2/2

Q62 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/2/1

Find the sum of first 20 terms of an A.P. whose $n^{t h}$ term is given as $a_{n} = 5 - 2 n$ .

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Answer:

S_{20} = - 320

$a_{1} = 5 - 2 = 3$ , $a_{20} = 5 - 40 = - 35$ .
$S_{20} = \frac{20}{2} (a_{1} + a_{20}) = 10 (3 - 35)$
$S_{20} = - 320$ .

Also asked in: 2022 Standard 30/2/2, 2022 Standard 30/2/3

Q32 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/2/3

In an A.P. if the sum of third and seventh term is zero, find its $5^{t h}$ term.

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Answer:

a_{5} = 0

$a_{3} + a_{7} = 0 \Rightarrow (a + 2 d) + (a + 6 d) = 0$
$2 a + 8 d = 0 \Rightarrow a + 4 d = 0$
$a_{5} = a + 4 d = 0$ .

Q3 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/2/3

Determine the A.P. whose third term is 5 and seventh term is 9.

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Answer: 3, 4, 5, 6, ...

$a + 2 d = 5$ and $a + 6 d = 9$ .
Subtracting: $4 d = 4 \Rightarrow d = 1$ ; then $a = 3$ .
The A.P. is 3, 4, 5, 6, ...

Q22 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/3/1

Find the sum of first 20 terms of an AP in which d = 5 and $a_{20}$ = 135.

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Answer: 1750

$a_{20} = a + 19 d \Rightarrow 135 = a + 95 \Rightarrow a = 40$ .
$S_{20} = \frac{20}{2} (a + a_{20}) = 10 (40 + 135) = 1750$ .

Also asked in: 2022 Standard 30/3/2

Q52 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/3/1

For what value of ‘n’, are the $n^{t h}$ terms of the APs : 9, 7, 5, ..... and 15, 12, 9, ..... the same ?

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Answer:

n = 7

First AP: $a = 9$ , $d = - 2$ , so $a_{n} = 9 - 2 (n - 1) = 11 - 2 n$ .
Second AP: $a = 15$ , $d = - 3$ , so $a_{n} = 15 - 3 (n - 1) = 18 - 3 n$ .
$11 - 2 n = 18 - 3 n \Rightarrow n = 7$ (both 7th terms equal $- 3$ ).

Also asked in: 2022 Standard 30/3/2, 2022 Standard 30/3/3

Q32 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/3/3

Find the common difference ‘d’ of an AP whose first term is 10 and the sum of the first 14 terms is 1505.

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Answer:

d = 15

$S_{14} = \frac{14}{2} [2 \times 10 + 13 d] = 7 (20 + 13 d)$ .
$7 (20 + 13 d) = 1505 \Rightarrow 20 + 13 d = 215 \Rightarrow 13 d = 195 \Rightarrow d = 15$ .

Q22 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/4/1

How many natural numbers are there between 1 and 1000 which are divisible by 5 but not by 2 ?

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Answer: 100

Numbers divisible by 5 but not by 2 are the odd multiples of 5: 5, 15, 25, ..., 995.
This is an A.P. with $a = 5$ , $d = 10$ , $a_{n} = 995$ .
$995 = 5 + (n - 1) \times 10 \Rightarrow n - 1 = 99 \Rightarrow n = 100$
There are 100 such numbers.

Q42 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/4/2

If the first term of an A.P. is 5, the last term is 15 and the sum of first $n$ terms is 30, then find the value of $n$ .

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Answer:

n = 3

$S_{n} = \frac{n}{2} (a + l)$
$30 = \frac{n}{2} (5 + 15) = 10 n$
$n = 3$

Q42 marksVery Short AnswerArithmetic Progressions CBSE 2022 · Standard 30/4/3

Find the sum of all 11 terms of an A.P. whose $6^{t h}$ term is 30.

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Answer: 330

$a_{6} = a + 5 d = 30$
$S_{11} = \frac{11}{2} (2 a + 10 d) = 11 (a + 5 d)$
$S_{11} = 11 \times 30 = 330$

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