CBSE Maths › Class 10 PYQs › Arithmetic Progressions

Arithmetic Progressions: 5 marks Questions (CBSE Class 10)

38 different 5 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)

Q325 marksLong AnswerArithmetic Progressions CBSE 2026 · Basic 430/4/1

The third and ninth term of an A.P. are 4 and – 8 respectively.
(i) Which term of the A.P. is zero ?
(ii) Find the value of n if $S_{n}$ = – 36.

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Answer: (i) 5th term (ii) n = 12

$a + 2 d = 4$ and $a + 8 d = - 8$ .
Subtracting: $6 d = - 12$ , so $d = - 2$ and $a = 8$ .
(i) $a_{n} = 8 + (n - 1) (- 2) = 10 - 2 n = 0 \Rightarrow n = 5$ . The 5th term is zero.
(ii) $S_{n} = \frac{n}{2} [16 + (n - 1) (- 2)] = n (9 - n)$ .
$n (9 - n) = - 36 \Rightarrow n^{2} - 9 n - 36 = 0 \Rightarrow (n - 12) (n + 3) = 0$ .
n = 12 (n cannot be negative).

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

Q335 marksLong AnswerArithmetic Progressions CBSE 2025 · Standard 30/3/1

The sum of the third term and the seventh term of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

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Answer: 76 (when

a = 1, d = \frac{1}{2}

) or 20 (when

a = 5, d = - \frac{1}{2}

)

$a_{3} + a_{7} = (a + 2 d) + (a + 6 d) = 2 a + 8 d = 6$ , so $a + 4 d = 3$ .
$a_{3} \times a_{7} = (a + 2 d) (a + 6 d) = 8$ . Put $a = 3 - 4 d$ : $(3 - 2 d) (3 + 2 d) = 8$ .
$9 - 4 d^{2} = 8$ , so $d^{2} = \frac{1}{4}$ , $d = \pm \frac{1}{2}$ .
If $d = \frac{1}{2}$ : $a = 1$ , $S_{16} = \frac{16}{2} [2 (1) + 15 \times \frac{1}{2}] = 8 \times \frac{19}{2} = 76$ .
If $d = - \frac{1}{2}$ : $a = 5$ , $S_{16} = 8 [10 - \frac{15}{2}] = 8 \times \frac{5}{2} = 20$ .

Also asked in: 2025 Standard 30/3/3

Q33 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2025 · Standard 30/3/1

The minimum age of children eligible to participate in a painting competition is 8 years. It is observed that the age of the youngest boy was 8 years and the ages of the participants, when seated in order of age, have a common difference of 4 months. If the sum of the ages of all the participants is 168 years, find the age of the eldest participant in the painting competition.

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Answer: 13 years

$a = 8$ years, $d = 4$ months $= \frac{1}{3}$ year, $S_{n} = 168$ .
$\frac{n}{2} [16 + (n - 1) \frac{1}{3}] = 168$ , so $n (48 + n - 1) = 1008$ .
$n^{2} + 47 n - 1008 = 0$ , so $(n + 63) (n - 16) = 0$ ; $n = 16$ ( $n > 0$ ).
Age of eldest $= a_{16} = 8 + 15 \times \frac{1}{3} = 13$ years.

Also asked in: 2025 Standard 30/3/3

Q325 marksLong AnswerArithmetic Progressions CBSE 2025 · Standard 30/3/2

An AP consists of 'n' terms whose n $^{th}$ term is 4 and the common difference is 2. If the sum of 'n' terms of AP is $- 14$ , then find 'n'. Also, find the sum of the first 20 terms.

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Answer:

n = 7

; sum of first 20 terms

= 220

$a_{n} = a + (n - 1) 2 = 4$ , so $a = 6 - 2 n$ .
$S_{n} = \frac{n}{2} (a + a_{n}) = \frac{n}{2} (6 - 2 n + 4) = n (5 - n) = - 14$ .
$n^{2} - 5 n - 14 = 0$ , so $(n - 7) (n + 2) = 0$ ; $n = 7$ ( $n > 0$ ).
$a = 6 - 14 = - 8$ , $d = 2$ .
$S_{20} = \frac{20}{2} [2 (- 8) + 19 (2)] = 10 (- 16 + 38) = 220$ .

Q32 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2025 · Standard 30/3/2

The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10 $^{th}$ term to the 30 $^{th}$ term is $1 : 3$ . Calculate the first and the thirteenth terms of the AP.

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Answer: First term

= 2

, thirteenth term

= 26

$S_{6} = \frac{6}{2} (2 a + 5 d) = 42$ , so $2 a + 5 d = 14$ ... (1)
$\frac{a + 9 d}{a + 29 d} = \frac{1}{3}$ , so $3 a + 27 d = a + 29 d$ , i.e. $a = d$ ... (2)
From (1) and (2): $7 a = 14$ , so $a = 2$ , $d = 2$ .
$a_{13} = a + 12 d = 2 + 24 = 26$ .

Q355 marksLong AnswerArithmetic Progressions CBSE 2024 · Basic 430/1/2

In an A.P. if $S_{n} = 4 n^{2} - n$ , then
(i) find the first term and common difference.
(ii) write the A.P.
(iii) which term of the A.P. is 107 ?

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Answer: (i)

a = 3

d = 8

(ii) 3, 11, 19, 27, ... (iii) 14th term

(i) $a_{1} = S_{1} = 4 - 1 = 3$ .
$S_{2} = 16 - 2 = 14$ , so $a_{2} = S_{2} - S_{1} = 11$ and $d = 11 - 3 = 8$ .
(ii) The A.P. is 3, 11, 19, 27, ...
(iii) $a_{n} = 3 + (n - 1) \times 8 = 8 n - 5$ .
$8 n - 5 = 107$ gives $n = 14$ , so 107 is the 14th term.

Q335 marksLong AnswerArithmetic Progressions CBSE 2024 · Basic 430/1/3

How many terms of the A.P. 27, 24, 21, ....... must be taken so that their sum is 105 ? Which term of the A.P. is zero ?

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Answer: 5 terms (or 14 terms); the 10th term is zero

$a = 27$ , $d = - 3$ .
$S_{n} = \frac{n}{2} [54 + (n - 1) (- 3)] = \frac{n}{2} (57 - 3 n) = 105$ .
$57 n - 3 n^{2} = 210$ , so $n^{2} - 19 n + 70 = 0$ , i.e. $(n - 5) (n - 14) = 0$ .
$n = 5$ or $n = 14$ ; both work, because terms 6th to 14th add up to 0.
$a_{n} = 27 + (n - 1) (- 3) = 30 - 3 n = 0$ gives $n = 10$ , so the 10th term is zero.

Q355 marksLong AnswerArithmetic Progressions CBSE 2024 · Basic 430/4/1

A manufacturer of TV sets produced 720 TV sets in the fourth year and 880 TV sets in the eighth year. Assuming that the production increases uniformly by a fixed number every year, find the production in the tenth year and the total production in the first seven years.

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Answer: Tenth year: 960 TV sets; first seven years: 5040 TV sets

Let the production form an A.P. with first term $a$ and common difference $d$ .
$a + 3 d = 720$ and $a + 7 d = 880$ .
Subtracting, $4 d = 160 \Rightarrow d = 40$ , so $a = 600$ .
$a_{10} = a + 9 d = 600 + 360 = 960$ .
$S_{7} = \frac{7}{2} (2 a + 6 d) = \frac{7}{2} (1200 + 240) = \frac{7}{2} \times 1440 = 5040$ .

Q335 marksLong AnswerArithmetic Progressions CBSE 2024 · Basic 430/4/2

The second term of an A.P. is 29 and the fourth term is 51. If the last term of the A.P. is 425, find how many terms are there and what is their sum.

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Answer: 38 terms; sum = 8417

$a + d = 29$ and $a + 3 d = 51 \Rightarrow 2 d = 22 \Rightarrow d = 11$ , $a = 18$ .
$a_{n} = 18 + (n - 1) 11 = 425 \Rightarrow (n - 1) 11 = 407 \Rightarrow n - 1 = 37 \Rightarrow n = 38$ .
$S_{38} = \frac{38}{2} (18 + 425) = 19 \times 443 = 8417$ .

Q345 marksLong AnswerArithmetic Progressions CBSE 2024 · Basic 430/4/3

If the sum of the first 7 terms of an A.P. is 91 and that of the first 17 terms is 561, then find the sum of the first n terms and hence find the $n^{t h}$ term.

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Answer:

S_{n} = 2 n^{2} - n

;

a_{n} = 4 n - 3

$S_{7} = \frac{7}{2} (2 a + 6 d) = 91 \Rightarrow a + 3 d = 13$ .
$S_{17} = \frac{17}{2} (2 a + 16 d) = 561 \Rightarrow a + 8 d = 33$ .
Subtracting, $5 d = 20 \Rightarrow d = 4$ , so $a = 1$ .
$S_{n} = \frac{n}{2} [2 (1) + (n - 1) 4] = \frac{n}{2} (4 n - 2) = 2 n^{2} - n$ .
$a_{n} = S_{n} - S_{n - 1} = (2 n^{2} - n) - (2 (n - 1)^{2} - (n - 1)) = 4 n - 3$ .

Q335 marksLong AnswerArithmetic Progressions CBSE 2024 · Standard 30/1/1

The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.

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Answer:

a = 2

d = 4

S_{20} = 800

; or

a = 30

d = - 4

S_{20} = - 160

Let the first term be $a$ and the eighth term $a_{8} = a + 7 d$ .
$a + a_{8} = 32$ and $a \cdot a_{8} = 60$ , so $a$ and $a_{8}$ are roots of $t^{2} - 32 t + 60 = 0$ , i.e. $(t - 2) (t - 30) = 0$ .
Case 1: $a = 2$ , $a_{8} = 30$ : $7 d = 28$ , $d = 4$ . $S_{20} = \frac{20}{2} [2 (2) + 19 (4)] = 10 \times 80 = 800$ .
Case 2: $a = 30$ , $a_{8} = 2$ : $7 d = - 28$ , $d = - 4$ . $S_{20} = 10 [60 + 19 (- 4)] = 10 \times (- 16) = - 160$ .

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

Q33 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2024 · Standard 30/1/1

In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.

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Answer: First term = 5, common difference = 3, sum of all 40 terms = 2540

$S_{9} = \frac{9}{2} (2 a + 8 d) = 153$ , so $a + 4 d = 17$ ... (1)
The last 6 terms are the 35th to 40th terms: sum $= \frac{6}{2} (a_{35} + a_{40}) = 3 (2 a + 73 d) = 687$ , so $2 a + 73 d = 229$ ... (2)
From (1), $a = 17 - 4 d$ . Substituting: $34 - 8 d + 73 d = 229$ , so $65 d = 195$ , $d = 3$ , $a = 5$ .
$S_{40} = \frac{40}{2} [2 (5) + 39 (3)] = 20 \times 127 = 2540$ .

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

Q355 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/2/1

If the sum of the first 7 terms of an A.P. is $- 14$ and that of 11 terms is $- 55$ , then find the sum of its first ‘n’ terms.

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Answer:

S_{n} = \frac{n ( 13 - 3 n )}{4}

$S_{7} = \frac{7}{2} (2 a + 6 d) = - 14$ , so $a + 3 d = - 2$ ...(1)
$S_{11} = \frac{11}{2} (2 a + 10 d) = - 55$ , so $a + 5 d = - 5$ ...(2)
(2) − (1): $2 d = - 3$ , $d = - \frac{3}{2}$ ; then $a = - 2 + \frac{9}{2} = \frac{5}{2}$
$S_{n} = \frac{n}{2} [5 + (n - 1) (- \frac{3}{2})] = \frac{n}{2} \cdot \frac{13 - 3 n}{2}$
$S_{n} = \frac{n ( 13 - 3 n )}{4}$

Q35 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/2/1

In an A.P., the sum of the first ‘n’ terms is $3 n^{2} + n$ . Find the first term and the common difference of the A.P. Hence, find its $1 5^{t h}$ term.

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Answer: First term = 4, common difference = 6,

a_{15} = 88

$S_{1} = 3 + 1 = 4$ , so a = 4
$S_{2} = 12 + 2 = 14$ , so $a_{2} = 14 - 4 = 10$
d = 10 − 4 = 6
$a_{15} = 4 + 14 \times 6 = 88$

Q335 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/2/2

If the sum of the first 7 terms of an A.P. is $- 21$ and that of the first 17 terms is $- 221$ , then find the sum of its first ‘n’ terms.

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Answer:

S_{n} = 4 n - n^{2}

$S_{7} = \frac{7}{2} (2 a + 6 d) = - 21$ , so $a + 3 d = - 3$ ...(1)
$S_{17} = \frac{17}{2} (2 a + 16 d) = - 221$ , so $a + 8 d = - 13$ ...(2)
(2) − (1): 5d = −10, d = −2; then a = −3 + 6 = 3
$S_{n} = \frac{n}{2} [6 + (n - 1) (- 2)] = \frac{n}{2} (8 - 2 n)$
$S_{n} = n (4 - n) = 4 n - n^{2}$

Q33 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/2/2

A man repays a loan of ₹ 3,250 by paying ₹ 20 in the first month and then increases the payment by ₹ 15 every month. How long will it take to clear the loan ?

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Answer: 20 months

The monthly payments form an A.P. with a = 20, d = 15, and $S_{n} = 3250$ .
$\frac{n}{2} [40 + (n - 1) 15] = 3250$
$n (15 n + 25) = 6500$ , so $3 n^{2} + 5 n - 1300 = 0$
$n = \frac{- 5 \pm 25 + 15600}{6} = \frac{- 5 \pm 125}{6}$
n = 20 (n cannot be negative), so the loan is cleared in 20 months.

Q335 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/2/3

The sum of the first 8 terms of an A.P. is 100 and the sum of its first 19 terms is 551. Find the sum of its first ‘n’ terms.

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Answer:

S_{n} = \frac{n ( 3 n + 1 )}{2}

$S_{8} = \frac{8}{2} (2 a + 7 d) = 100$ , so $2 a + 7 d = 25$ ...(1)
$S_{19} = \frac{19}{2} (2 a + 18 d) = 551$ , so $a + 9 d = 29$ , i.e. $2 a + 18 d = 58$ ...(2)
(2) − (1): 11d = 33, d = 3; then 2a = 25 − 21 = 4, a = 2
$S_{n} = \frac{n}{2} [4 + (n - 1) 3] = \frac{n ( 3 n + 1 )}{2}$

Q33 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/2/3

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

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Answer: Proved.

Let the first term be a and the common difference d.
$S_{p} = S_{q}$ : $\frac{p}{2} [2 a + (p - 1) d] = \frac{q}{2} [2 a + (q - 1) d]$
$2 a (p - q) + d [(p^{2} - q^{2}) - (p - q)] = 0$
$(p - q) [2 a + (p + q - 1) d] = 0$
Since $p \neq = q$ , $2 a + (p + q - 1) d = 0$
$S_{p + q} = \frac{p + q}{2} [2 a + (p + q - 1) d] = \frac{p + q}{2} \times 0 = 0$

Q335 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/4/1

Find the sum of first 25 terms of the A.P. whose $n^{t h}$ term is given by $a_{n} = 5 + 6 n$ . Also, find the ratio of $2 0^{t h}$ term to $4 5^{t h}$ term.

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Answer:

S_{25} = 2075

;

a_{20} : a_{45} = 5 : 11

$a_{1} = 5 + 6 = 11$ , $a_{25} = 5 + 150 = 155$ .
$S_{25} = \frac{25}{2} (11 + 155) = \frac{25}{2} \times 166 = 2075$ .
$a_{20} = 5 + 120 = 125$ , $a_{45} = 5 + 270 = 275$ .
$a_{20} : a_{45} = 125 : 275 = 5 : 11$ .

Q33 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/4/1

In an A.P., if $S_{n} = 3 n^{2} + 5 n$ and $a_{k} = 164$ , find the value of k.

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Answer:

k = 27

$a_{n} = S_{n} - S_{n - 1} = 3 n^{2} + 5 n - [3 (n - 1)^{2} + 5 (n - 1)]$ .
$= 3 (2 n - 1) + 5 = 6 n + 2$ .
$a_{k} = 6 k + 2 = 164 \Rightarrow 6 k = 162 \Rightarrow k = 27$ .

Q325 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/4/2

Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18, respectively.

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Answer: 5610

$d = 18 - 14 = 4$ , $a = 14 - 4 = 10$ .
$S_{51} = \frac{51}{2} [2 (10) + 50 (4)] = \frac{51}{2} \times 220$ .
$= 51 \times 110 = 5610$ .

Q32 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/4/2

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

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Answer:

n = 16

d = \frac{8}{3}

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

Q345 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/4/3

The first term of an A.P. is $- 5$ and the last term is 45. If the sum of all the terms of the A.P. is 120, find the number of terms and the common difference.

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Answer:

n = 6

d = 10

$S_{n} = \frac{n}{2} (a + l) \Rightarrow 120 = \frac{n}{2} (- 5 + 45) = 20 n \Rightarrow n = 6$ .
$l = a + (n - 1) d \Rightarrow 45 = - 5 + 5 d$ .
$d = 10$ .

Q34 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/4/3

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 first $n$ terms.

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Answer:

S_{n} = n^{2}

$S_{7} = \frac{7}{2} (2 a + 6 d) = 49 \Rightarrow a + 3 d = 7$ ... (1)
$S_{17} = \frac{17}{2} (2 a + 16 d) = 289 \Rightarrow a + 8 d = 17$ ... (2)
Subtracting: $5 d = 10 \Rightarrow d = 2$ , and $a = 1$ .
$S_{n} = \frac{n}{2} [2 (1) + (n - 1) 2] = \frac{n}{2} \times 2 n = n^{2}$ .

Q335 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/6/1

The first term of an A.P. is 22, the last term is $- 6$ and the sum of all the terms is 64. Find the number of terms of the A.P. Also, find the common difference.

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Answer: Number of terms = 8; common difference =

- 4

$S_{n} = \frac{n}{2} (a + l)$ : $64 = \frac{n}{2} (22 - 6) = 8 n$ , so $n = 8$ .
$l = a + (n - 1) d$ : $- 6 = 22 + 7 d$ , so $7 d = - 28$ and $d = - 4$ .

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

Q345 marksLong AnswerArithmetic Progressions CBSE 2023 · Basic 430/6/3

The sum of the $4^{th}$ and $8^{th}$ term of an A.P. is 24 and the sum of the $6^{th}$ and $1 0^{th}$ term of the A.P. is 44. Find the A.P. Also, find the sum of first 25 terms of the A.P.

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Answer: A.P.:

- 13, - 8, - 3, 2, \dots

;

S_{25} = 1175

$a_{4} + a_{8} = 24$ : $(a + 3 d) + (a + 7 d) = 24$ , so $a + 5 d = 12$ .
$a_{6} + a_{10} = 44$ : $(a + 5 d) + (a + 9 d) = 44$ , so $a + 7 d = 22$ .
Subtracting: $2 d = 10$ , so $d = 5$ and $a = 12 - 25 = - 13$ .
A.P.: $- 13, - 8, - 3, 2, \dots$
$S_{25} = \frac{25}{2} [2 (- 13) + 24 \times 5] = \frac{25}{2} \times 94 = 1175$ .

Q355 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/1/1

The ratio of the 11 $^{th}$ term to 17 $^{th}$ term of an A.P. is 3 : 4. Find the ratio of 5 $^{th}$ term to 21 $^{st}$ term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.

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Answer: 3 : 7 and 25 : 189

$\frac{a + 10 d}{a + 16 d} = \frac{3}{4}$ , so $4 a + 40 d = 3 a + 48 d$ , $a = 8 d$ .
$\frac{a _{5}}{a _{21}} = \frac{a + 4 d}{a + 20 d} = \frac{12 d}{28 d} = \frac{3}{7}$ .
$S_{5} = \frac{5}{2} (2 a + 4 d) = \frac{5}{2} (20 d) = 50 d$ .
$S_{21} = \frac{21}{2} (2 a + 20 d) = \frac{21}{2} (36 d) = 378 d$ .
$S_{5} : S_{21} = 50 : 378 = 25 : 189$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q35 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/1/1

250 logs are stacked in the following manner :
22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row ?

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Answer: 20 rows; 3 logs in the top row

Rows form an A.P.: a = 22, d = –1.
$S_{n} = \frac{n}{2} [44 + (n - 1) (- 1)] = 250$ , so $n (45 - n) = 500$ .
$n^{2} - 45 n + 500 = 0$ , $(n - 20) (n - 25) = 0$ , n = 20 or 25.
For n = 25 the last row would have 22 – 24 = –2 logs, not possible. So n = 20.
Top row $a_{20} = 22 + 19 (- 1) = 3$ logs.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q325 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/2/1

How many terms of the arithmetic progression 45, 39, 33, ........ must be taken so that their sum is 180 ? Explain the double answer.

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Answer: 6 terms or 10 terms (the 7th to 10th terms 9, 3,

- 3

- 9

add up to 0).

$a = 45$ , $d = - 6$
$S_{n} = \frac{n}{2} [90 + (n - 1) (- 6)] = \frac{n}{2} (96 - 6 n) = 48 n - 3 n^{2}$
$48 n - 3 n^{2} = 180 \Rightarrow n^{2} - 16 n + 60 = 0 \Rightarrow (n - 6) (n - 10) = 0$
$n = 6$ or $n = 10$
The 7th to 10th terms are 9, 3, $- 3$ , $- 9$ , whose sum is 0.
So the sum of the first 6 terms and of the first 10 terms is the same, 180.

Q345 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/2/2

Solve the equation for x :
$1 + 4 + 7 + 10 + .... + x = 287$

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Answer:

x = 40

The left side is an AP with $a = 1$ , $d = 3$ . Let it have $n$ terms.
$S_{n} = \frac{n}{2} [2 + (n - 1) 3] = \frac{n ( 3 n - 1 )}{2} = 287$
$3 n^{2} - n - 574 = 0 \Rightarrow (n - 14) (3 n + 41) = 0$
$n = 14$ (n must be a natural number)
$x = a_{14} = 1 + 13 \times 3 = 40$

Q335 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/2/3

Prerna saves ₹ 32 during the first month, ₹ 36 in the second month and ₹ 40 in the third month. If she continues to save in this manner, in how many months will she save ₹ 2,000 ?

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Answer: 25 months

The savings form an AP with $a = 32$ , $d = 4$ .
$S_{n} = \frac{n}{2} [64 + (n - 1) 4] = n (30 + 2 n) = 2000$
$n^{2} + 15 n - 1000 = 0 \Rightarrow (n + 40) (n - 25) = 0$
$n = 25$ (n cannot be negative)

Q345 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/6/1

The ratio of the 11 $^{th}$ term to the 18 $^{th}$ term of an A.P. is 2 : 3. Find the ratio of the 5 $^{th}$ term to the 21 $^{st}$ term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms.

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Answer: 1 : 3 and 5 : 49

$\frac{a + 10 d}{a + 17 d} = \frac{2}{3}$
$3 a + 30 d = 2 a + 34 d$ , so $a = 4 d$ .
$\frac{a _{5}}{a _{21}} = \frac{a + 4 d}{a + 20 d} = \frac{8 d}{24 d} = \frac{1}{3}$
$S_{5} = \frac{5}{2} (2 a + 4 d) = \frac{5}{2} (12 d) = 30 d$
$S_{21} = \frac{21}{2} (2 a + 20 d) = \frac{21}{2} (28 d) = 294 d$
$S_{5} : S_{21} = 30 : 294 = 5 : 49$

Q34 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/6/1

If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

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Answer: 100

$S_{6} = 3 (2 a + 5 d) = 36$ , so $2 a + 5 d = 12$ .
$S_{16} = 8 (2 a + 15 d) = 256$ , so $2 a + 15 d = 32$ .
Subtracting: $10 d = 20$ , d = 2, then a = 1.
$S_{10} = 5 (2 a + 9 d) = 5 (2 + 18) = 100$

Q355 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/6/2

Find the sum of integers between 100 and 200 which are (i) divisible by 9 (ii) not divisible by 9.

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Answer: (i) 1683 (ii) 13167

(i) Integers divisible by 9: 108, 117, ..., 198.
$198 = 108 + (n - 1) 9$ gives n = 11.
Sum $= \frac{11}{2} (108 + 198) = 11 \times 153 = 1683$
(ii) Sum of all integers from 101 to 199: 99 terms, sum $= \frac{99}{2} (101 + 199) = 14850$ .
Sum of those not divisible by 9 = 14850 – 1683 = 13167

Q35 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/6/2

Solve the equation :
$- 4 + (- 1) + 2 + 5 + \dots + x = 437$ .

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Answer: x = 50

The terms form an A.P. with a = –4, d = 3.
$S_{n} = \frac{n}{2} [2 (- 4) + (n - 1) 3] = \frac{n}{2} (3 n - 11) = 437$
$3 n^{2} - 11 n - 874 = 0$
$D = 121 + 10488 = 10609 = 10 3^{2}$ , so $n = \frac{11 + 103}{6} = 19$ (the negative root is rejected).
$x = a_{19} = - 4 + 18 \times 3 = 50$

Q325 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/6/3

The sum of first seven terms of an A.P. is 182. If its 4 $^{th}$ term and the 17 $^{th}$ term are in the ratio 1 : 5, find the A.P.

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Answer: 2, 10, 18, 26, ...

$S_{7} = \frac{7}{2} (2 a + 6 d) = 7 (a + 3 d) = 182$ , so $a + 3 d = 26$ .
$\frac{a + 3 d}{a + 16 d} = \frac{1}{5}$ , so $a + 16 d = 5 \times 26 = 130$ .
Subtracting: $13 d = 104$ , d = 8; then a = 26 – 24 = 2.
The A.P. is 2, 10, 18, 26, ...

Q32 (OR) (OR)5 marksLong AnswerArithmetic Progressions CBSE 2023 · Standard 30/6/3

The sum of first q terms of an A.P. is $63 q - 3 q^{2}$ . If its p $^{th}$ term is –60, find the value of p. Also, find the 11 $^{th}$ term of this A.P.

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Answer: p = 21; 11th term = 0

$a_{1} = S_{1} = 63 - 3 = 60$ ; $S_{2} = 126 - 12 = 114$ , so $a_{2} = 54$ and d = –6.
$a_{p} = 60 + (p - 1) (- 6) = 66 - 6 p$
$66 - 6 p = - 60$ , so p = 21.
$a_{11} = 66 - 66 = 0$

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