Arithmetic Progressions — Class 10 Maths PYQs

107 previous-year board questions (2023–2025) with marking-scheme solutions, grouped by topic and marks.

Try each question first, then press (or tap Show Solution) to reveal the answer. Press again for the next question.

Basics of AP, CD, Make, find x

1 Mark Questions
11 Mark · July 2023 · Standardopen ↗
If $x, 2x + 9, 4x + 3$ are three consecutive terms of an A.P., then the value of $x$ is:
  • (a)$3$
  • (b)$13$
  • (c)$10$
  • (d)$15$
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(d) $15$
21 Mark · July 2023 · Standardopen ↗
If $2x$, $x + 10$, $3x + 2$ are three consecutive terms of an A.P., then the value of $x$ is:
  • (a)$4$
  • (b)(e) $6$
  • (c)$5$
  • (d)$8$
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Sol. (c) $6$
31 Mark · July 2023 · Standardopen ↗
If $x + 1$, $3x$ and $4x + 2$ are three consecutive terms of an A.P., then the value of $x$ is:
  • (a)$2$
  • (b)$4$
  • (c)$3$
  • (d)$5$
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(b) $3$
41 Mark · March 2023 · Standardopen ↗
If $p-1$, $p + 1$ and $2p + 3$ are in A.P., then the value of $p$ is
  • (a)-2
  • (b)4
  • (c)0
  • (d)2
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(C) 0
51 Mark · March 2023 · Standardopen ↗
If $a, b, c$ form a A.P. with common difference $d$, then the value of $a - 2b - c$ is equal to
  • (a)$2a + 4d$
  • (b)0
  • (c)$-2a - 4d$
  • (d)$2a - 3d$
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(C) $-2a - 4d$
61 Mark · March 2023 · Standardopen ↗
If $p-1, p + 1$ and $2p + 3$ are in A.P., then the value of $p$ is
  • (a)$-2$
  • (b)$4$
  • (c)$0$
  • (d)$2$
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(C) $0$
71 Mark · March 2023 · Standardopen ↗
The next term of the A.P. : $\sqrt{6}$, $\sqrt{24}$, $\sqrt{54}$ is :
  • (a)$\sqrt{60}$
  • (b)$\sqrt{96}$
  • (c)$\sqrt{72}$
  • (d)$\sqrt{216}$
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Sol. (b) $\sqrt{96}$
81 Mark · March 2023 · Standardopen ↗
If $k + 2$, $4k - 6$ and $3k - 2$ are three consecutive terms of an A.P., then the value of $k$ is :
  • (a)$3$
  • (b)$-3$
  • (c)$4$
  • (d)$-4$
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(a) $3$
91 Mark · March 2023 · Standardopen ↗
The next term of the A.P. : $\sqrt{7}$, $\sqrt{28}$, $\sqrt{63}$ is :
  • (a)$\sqrt{70}$
  • (b)$\sqrt{80}$
  • (c)$\sqrt{97}$
  • (d)$\sqrt{112}$
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(d) $\sqrt{112}$
101 Mark · March 2023 · Standardopen ↗
The common difference of the A.P. whose $n^{\text{th}}$ term is given by $a_n = 3n + 7$, is:
  • (a)$7$
  • (b)$3$
  • (c)$3n$
  • (d)$1$
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(b) $3$
111 Mark · March 2024 · Standardopen ↗
If $k +7$, $2k - 2$ and $2k + 6$ are three consecutive terms of an A.P., then the value of $k$ is:
  • (a)$15$
  • (b)$5$
  • (c)$17$
  • (d)$1$
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(B) $17$
121 Mark · March 2024 · Standardopen ↗
$n^{th}$ term of an A.P. is $7n + 4$. The common difference is :
  • (a)$7n$
  • (b)$4$
  • (c)$7$
  • (d)$1$
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(c) $7$
131 Mark · March 2024 · Standardopen ↗
The common difference of an A.P. in which $a_{15} - a_{11} = 48$, is
  • (a)$12$
  • (b)$-12$
  • (c)$16$
  • (d)$-16$
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(A) $12$
141 Mark · March 2024 · Standardopen ↗
The common difference of an A.P. in which $a_{20} - a_{15} = 20$, is
  • (a)4
  • (b)5
  • (c)$4d$
  • (d)$5d$
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(A) 4
151 Mark · March 2024 · Standardopen ↗
The common difference of the A.P. $\frac{1}{2x}$, $\frac{1-4x}{2x}$, $\frac{1-8x}{2x}$
dots is :
  • (a)$-2x$
  • (b)$-2$
  • (c)$2$
  • (d)$2x$
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(B) $-2$
161 Mark · March 2024 · Standardopen ↗
The common difference of the A.P.
$\frac{1-4x}{2x}$, $\frac{1-4x}{2x}$, $\frac{1-8x}{2x}$ .......is :
  • (a)$-2x$
  • (b)$2$
  • (c)$-2$
  • (d)$2x$
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Sol.
(B) $-2$
171 Mark · March 2024 · Standardopen ↗
If the first three terms of an A.P. are $3p - 1, 3p + 5, 5p + 1$ respectively; then the value of $p$ is :
  • (a)$2$
  • (b)$-3$
  • (c)$4$
  • (d)$5$
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(D) $5$
181 Mark · March 2024 · Standardopen ↗
The next ($4^{\text{th}}$) term of the A.P. $\sqrt{18}$, $\sqrt{50}$, $\sqrt{98}$, .. is:
  • (a)$\sqrt{128}$
  • (b)$\sqrt{140}$
  • (c)$\sqrt{162}$
  • (d)$\sqrt{200}$
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(C) $\sqrt{162}$
191 Mark · March 2024 · Standardopen ↗
The next ($4^{\text{th}}$) term of the A.P. $\sqrt{7}$, $\sqrt{28}$, $\sqrt{63}$, ... is:
  • (a)$\sqrt{70}$
  • (b)$\sqrt{97}$
  • (c)$\sqrt{84}$
  • (d)$\sqrt{112}$
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(D) $\sqrt{112}$
201 Mark · July 2025 · Standardopen ↗
The $6^{th}$ term of the AP $\sqrt{27}$, $\sqrt{75}$, $\sqrt{147}$, ... is :
  • (a)$\sqrt{243}$
  • (b)$\sqrt{363}$
  • (c)$\sqrt{300}$
  • (d)$\sqrt{507}$
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(D) $\sqrt{507}$
211 Mark · March 2025 · Standardopen ↗
Assertion (A): Common difference of the AP: $5, 1, - 3, - 7, \dots$ is $4$. Reason (R): Common difference of the AP : $a_1, a_2, a_3, \dots, a_n$ is obtained by $d = a_n - a_{n-1}$
  • (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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(D) Assertion (A) is false, but Reason (R) is true.
221 Mark · March 2025 · Standardopen ↗
Questions number $19$ and $20$ are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.
(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
textbf{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.
Assertion (A): Common difference of the AP : $5, 1, -3, -7, ...$ is $4$.
Reason (R): Common difference of the AP : $a_1, a_2, a_3, ..., a_n$ is obtained by $d = a_n - a_{n-1}$
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(D) Assertion (A) is false, but Reason (R) is true.

Term Formula

1 Mark Questions
231 Mark · July 2023 · Standardopen ↗
The $4^{th}$ term from the end of an AP $-11, -8, -5, \ldots, 49$ is :
  • (a)$40$
  • (b)$37$
  • (c)$43$
  • (d)$58$
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Ans. (a) $40$
241 Mark · March 2023 · Standardopen ↗
The $11^{\text{th}}$ term from the end of the A.P.: $10, 7, 4, ......., - 62$ is:
  • (a)$25$
  • (b)$16$
  • (c)$-32$
  • (d)$0$
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(c) $-32$
251 Mark · March 2023 · Standardopen ↗
The $13^{th}$ term from the end of the A.P. : $20, 13, 6, -1, ....., -148$ is :
  • (a)$57$
  • (b)$-57$
  • (c)$64$
  • (d)$-64$
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(d) $-64$
261 Mark · March 2023 · Standardopen ↗
The $11^{\text{th}}$ term from the end of the A.P. : $10, 7, 4, \dots, -62$ is:
  • (a)$25$
  • (b)$-32$
  • (c)$16$
  • (d)$0$
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(c) $-32$
271 Mark · March 2024 · Standardopen ↗
The $7^{th}$ term from the end of the A.P. : $-8,-5,- 2, ..., 49$ is :
  • (a)$67$
  • (b)$31$
  • (c)$13$
  • (d)$10$
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(C) $31$
281 Mark · July 2024 · Standardopen ↗
If $5$ times the $5^{th}$ term of an A.P. is equal to $9$ times the $9^{th}$ term, then its $14^{th}$ term is :
  • (a)$5$
  • (b)$9$
  • (c)$0$
  • (d)$14$
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Sol. (C) $0$
291 Mark · July 2024 · Standardopen ↗
Assertion (A) : The eighth term of the A.P. $\frac{1}{m}$, $\frac{1+2m}{m}$, $\frac{1+4m}{m}$,
dots is $\frac{1+14m}{m}$.
Reason (R) : The $n^{th}$ term of A.P. ($a_n$) = $a + (n - 1) d$.
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Sol. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
301 Mark · March 2024 · Standardopen ↗
Which term of the A.P. $-29, -26, -23, \dots, 61$ is $16$?
  • (a)$11^{th}$
  • (b)$10^{th}$
  • (c)$16^{th}$
  • (d)$31^{st}$
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(B) $16^{th}$
311 Mark · March 2024 · Standardopen ↗
Which term of the A.P. -29, -26, -23, ....., 61 is 16?
  • (a)11th
  • (b)16th
  • (c)10th
  • (d)31st
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(B) 16th
321 Mark · March 2024 · Standardopen ↗
The number of terms in the A.P. $3, 6, 9, 12, ..., 111$ is:
  • (a)$36$
  • (b)$40$
  • (c)$37$
  • (d)$30$
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(C) $37$
331 Mark · March 2024 · Standardopen ↗
The $14^{th}$ term from the end of the A.P. $-11$, $-8$, $-5$, ..., $49$ is:
  • (a)$7$
  • (b)$13$
  • (c)$10$
  • (d)$28$
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(B) $10$
341 Mark · July 2025 · Standardopen ↗
Which term of the AP 7, 10, 13, ... is 52?
  • (a)$13^{th}$
  • (b)$16^{th}$
  • (c)$15^{th}$
  • (d)$17^{th}$
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(C) $16^{th}$
351 Mark · July 2025 · Standardopen ↗
If the $23^{rd}$ term of an AP exceeds its $16^{th}$ term by $21$, then the common difference is :
  • (a)$1$
  • (b)$2$
  • (c)$3$
  • (d)$7$
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(C) $3$
361 Mark · July 2025 · Standardopen ↗
The $21^{st}$ term of an AP, whose first two terms are $-3$ and $4$ respectively, is :
  • (a)$17$
  • (b)$143$
  • (c)$137$
  • (d)$153$
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(C) $137$
371 Mark · July 2025 · Standardopen ↗
The $5^{th}$ term from the end of an AP $-11, -8, -5,..., 55$ is :
  • (a)$1$
  • (b)$43$
  • (c)$40$
  • (d)$46$
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(B) $43$
381 Mark · March 2025 · Standardopen ↗
The $11^{\text{th}}$ and $13^{\text{th}}$ term of an AP are $39$ and $45$, respectively. What is the common difference of the AP?
  • (a)$42$
  • (b)$21$
  • (c)$6$
  • (d)$3$
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(D) $3$
391 Mark · March 2025 · Standardopen ↗
The $10^{th}$ term of the AP $5, \frac{19}{4}, \frac{9}{2}, \frac{17}{4}, \dots$ is:
  • (a)$\frac{11}{4}$
  • (b)$\frac{4}{11}$
  • (c)$\frac{13}{4}$
  • (d)$\frac{4}{13}$
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(A) $\frac{11}{4}$
401 Mark · March 2025 · Standardopen ↗
The $9^{th}$ term from the end (towards first term) of the AP $7, 11, 15, 19, ..., 147$ is:
  • (a)$135$
  • (b)$125$
  • (c)$115$
  • (d)$39$
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(C) $115$
411 Mark · March 2025 · Standardopen ↗
Assertion (A): For an A.P., 3, 6, 9, ..., 198, $10^{th}$ term from the end is 168. Reason (R): If 'a' and 'l' are the first term and last term of an A.P. with common difference 'd', then $n^{th}$ term from the end of the given A.P. is $l - (n-1)d$.
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(D) Assertion (A) is false, but Reason (R) is true.
421 Mark · March 2025 · Standardopen ↗
Directions : In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from following : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
Assertion (A) : For an A.P., 3, 6, 9, ..., 198, $10^{th}$ term from the end is 168.
Reason (R) : If 'a' and 'l' are the first term and last term of an A.P. with common difference 'd', then $n^{th}$ term from the end of the given A.P. is $l - (n - 1)d$.
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(D) Assertion (A) is false, but Reason (R) is true.
3 Marks Questions
433 Marks · March 2023 · Standardopen ↗
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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Sol. $a = -14, a_5 = 2 \Rightarrow a+4d=2$
$-14 + 4d = 2 \Rightarrow d = 4$
$a_n = 62 \Rightarrow a + (n-1)d = 62$
$-14 + (n - 1)4 = 62 \Rightarrow n = 20$
443 Marks · March 2023 · Standardopen ↗
Which term of the A.P. : $65, 61, 57, 53, \dots$ is the first negative term ?
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Sol. $65, 61, 57, 53, \dots$
$a = 65, d = -4$
Let $a_n$ be the first negative term
$a_n < 0 \Rightarrow a+(n-1)d < 0$
$65 + (n - 1) (-4) <0 \Rightarrow 69 – 4n < 0$
$n > \frac{69}{4}$
$\therefore$ Least positive integral value of $n$ which satisfies $n > \frac{69}{4}$ is $18$
$\therefore 1^{st}$ negative term of the AP = $18$
453 Marks · July 2025 · Standardopen ↗
The sum of the third and seventh terms of an AP is $40$ and the sum of its sixth and fourteenth terms is $70$. Find the sum of the first ten terms of the AP.
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$a_3 + a_7 = 40$
$(a + 2d) + (a + 6d) = 40$
$\Rightarrow 2a + 8d = 40$ or $a + 4d = 20$ --- (1)
$a_6 + a_{14} = 70$
$(a + 5d) + (a + 13d) = 70$
$\Rightarrow 2a +18d = 70$ or $a + 9d = 35$ --- (2)
Solving (1) and (2), we get
$a = 8$ and $d = 3$
$S_{10} = \frac{10}{2} \times [2 (8) + 9 (3)]$
$= 215$

Sum Formula

1 Mark Questions
461 Mark · March 2023 · Standardopen ↗
DIRECTIONS : In the question number $19$ and $20$, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option out of the following :
Assertion (A) : $a, b, c$ are in A.P. if and only if $2b = a + c$.
Reason (R) : The sum of first $n$ odd natural numbers is $n^2$.
  • (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • (b)Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of 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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(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
471 Mark · March 2024 · Standardopen ↗
The sum of first $200$ natural numbers is
  • (a)$2010$
  • (b)$20100$
  • (c)$2000$
  • (d)$21000$
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(C) $20100$
481 Mark · March 2024 · Standardopen ↗
If the sum of first $n$ terms of an A.P. is $3n^2 + 4n$ and its common difference is $6$, then its first term is :
  • (a)$7$
  • (b)$6$
  • (c)$4$
  • (d)$3$
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(A) $7$
491 Mark · March 2024 · Standardopen ↗
The sum of the first three terms of an AP is $30$ and the sum of the last three terms is $36$. If the first term is $9$, then the number of terms is :
  • (a)$10$
  • (b)$5$
  • (c)$6$
  • (d)$13$
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(B) $5$
501 Mark · March 2024 · Standardopen ↗
Assertion (A): The sum of the first fifteen terms of the AP $21, 18, 15, 12, \dots$ is zero.
Reason (R) : The sum of the first $n$ terms of an AP with first term '$a$' and common difference '$d$' is given by $S_n = \frac{n}{2} [a + (n - 1) d]$.
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(C) Assertion (A) is true, but Reason (R) is false.
511 Mark · July 2025 · Standardopen ↗
If the sum of the first 'n' terms of an AP is $2n^2 + 5n$, then its common difference is :
  • (a)$2$
  • (b)$4$
  • (c)$5$
  • (d)$7$
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(B) $4$
521 Mark · March 2025 · Standardopen ↗
If the sum of first $m$ terms of an $AP$ is $2m^2 + 3m$, then its second term is:
  • (a)$10$
  • (b)$9$
  • (c)$12$
  • (d)$4$
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(B) $9$
531 Mark · March 2025 · Standardopen ↗
If the first term of an A.P. is $-12$ and the common difference is $4$, then the sum of its first $7$ terms is
  • (a)$-24$
  • (b)$-48$
  • (c)$0$
  • (d)$48$
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(C) $0$
3 Marks Questions
543 Marks · March 2024 · Standardopen ↗
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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Let $a$ be the first term and $d$ be the common difference.
$\frac{7}{2}(2a + 6d) = 49$
$a+3d=7 \dots (i)$
$\frac{17}{2}(2a + 16d) = 289$
$a + 8d = 17 \dots (ii)$
solving (i) and (ii)
$d=2 \& a=1$
$S_{20} = \frac{20}{2} [2(1)+19(2)]$
$= 400$
5 Marks Questions
555 Marks · March 2023 · Standardopen ↗
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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$45, 39, 33, .......$
a = $45$, d = $- 6$
Sn = $180$
$180 = \frac{n}{2} [2 \times 45 + (n - 1) (-6)]$
$180 = \frac{n}{2} [90-6n + 6]$
$\Rightarrow 360 = 96n - 6n^2$
$\Rightarrow 6n^2 - 96n + 360 = 0$
$\Rightarrow n^2 - 16n + 60 = 0 \Rightarrow (n - 10) (n - 6) = 0$
n - 10 = $0$, n - 6 = $0 \Rightarrow n = 10, 6$
We get two values of 'n' as sum of $7^{th}$ term to $10^{th}$ term is zero as some terms are negative and some are positive.
565 Marks · March 2023 · Standardopen ↗
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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$S_6 = 36 \Rightarrow \frac{6}{2}[2a + 5d] = 36$
$2a + 5d = 12$ ----- (1)
$S_{16} = 256 \Rightarrow \frac{16}{2}[2a + 15d] = 256$
$2a +15d = 32$ ----- (2)
Solving (1) and (2)
$d = 2$
$a = 1$
$S_{10} = \frac{10}{2}[2(1) +9(2)]$
$= 100$

Term & Sum Mix

1 Mark Questions
571 Mark · March 2023 · Standardopen ↗
Assertion (A) : $a, b, c$ are in A.P. if and only if $2b = a + c$.
Reason (R) : The sum of first $n$ odd natural numbers is $n^2$.
  • (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • (b)Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of 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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Sol. (b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
581 Mark · March 2023 · Standardopen ↗
If the sum of the first $n$ terms of an A.P be $3n^2 + n$ and its common difference is $6$, then its first term is
  • (a)2
  • (b)3
  • (c)1
  • (d)4
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(D) 4
591 Mark · July 2024 · Standardopen ↗
In an A.P., the first and last terms are $7$ and $73$ respectively. If the sum of all its terms is $480$, then the number of terms of the A.P. is:
  • (a)$6$
  • (b)$12$
  • (c)$18$
  • (d)$30$
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(B) $12$
601 Mark · March 2024 · Standardopen ↗
In an A.P., if the first term $a = 7$, $n$th term $a_n = 84$ and the sum of first $n$ terms $s_n = \frac{2093}{2}$, then $n$ is equal to :
  • (a)$22$
  • (b)$24$
  • (c)$23$
  • (d)$26$
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Sol. (c) $23$
611 Mark · March 2024 · Standardopen ↗
In an A.P., if the first term $(a) = -16$ and the common difference $(d) = -2$, then the sum of first 10 terms is :
  • (a)$-200$
  • (b)$-70$
  • (c)$-250$
  • (d)$250$
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(c) $-250$
3 Marks Questions
623 Marks · March 2023 · Standardopen ↗
The sum of first $15$ terms of an A.P. is $750$ and its first term is $15$. Find its $20^{th}$ term.
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$a = 15, S_{15} = 750$
$\frac{15}{2}[2a + 14d] = 750$
$2(15) + 14d = 100$
$d = 5$
$a_{20} = a + 19d = 15 + 19(5) = 110$
$\text{OR}$
633 Marks · March 2023 · Standardopen ↗
If $p^{\text{th}}$ term of an A.P. is $q$ and $q^{\text{th}}$ term is $p$, then prove that its $n^{\text{th}}$ term is $(p + q - n)$.
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$a_p = a + (p - 1)d = q$
quad (i)
$a_q = a + (q - 1)d = p$
quad (ii)
Solving (i) and (ii)
$d = -1, a = q + p - 1$
$a_n = (q + p - 1) + (n - 1)(-1) = q + p - n$
643 Marks · March 2023 · Standardopen ↗
In an A.P., the sum of the first $n$ terms is given by $S_n = 6n - n^2$. Find its $30^{th}$ term.
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Here $S_n = 6n - n^2$
For $n = 1$, $S_1 = a_1 = 6(1) - (1)^2 = 6 - 1 = 5$
So, the first term $a = 5$.
For $n = 2$, $S_2 = a_1 + a_2 = 6(2) - (2)^2 = 12 - 4 = 8$
We know $S_2 = a_1 + a_2 = a + (a+d) = 2a+d$.
So, $2a+d = 8$.
Substitute $a=5$: $2(5) + d = 8$
$10 + d = 8$
$d = 8 - 10 = -2$
Now, find the $30^{th}$ term $a_{30}$:
$a_{30} = a + (30-1)d$
$a_{30} = 5 + 29(-2)$
$a_{30} = 5 - 58$
$a_{30} = -53$
653 Marks · March 2023 · Standardopen ↗
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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$a = 8, l = 65$
$$\begin{aligned}& 730 = \frac{n}{2}[8+65] \\ & n = \frac{730 \times 2}{73} = 20\end{aligned}$$
$$\begin{aligned}& \therefore l = a + 19d \implies 65 = 8 + 19d \\ & \implies d = 3\end{aligned}$$
663 Marks · March 2024 · Standardopen ↗
If the sum of first $m$ terms of an A.P. is same as sum of its first $n$ terms ($m \ne n$), then show that the sum of its first $(m + n)$ terms is zero.
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$S_m = S_n$
$\Rightarrow \frac{m}{2}[2a + (m - 1)d] = \frac{n}{2}[2a + (n - 1)d]$
$\Rightarrow 2a(m - n) = d(n^2 - m^2) - d(n - m)$
$\Rightarrow 2a = -d(m + n - 1)$
or $2a + (m + n - 1)d = 0$
i.e., $S_{m+n} = \frac{m+n}{2}[2a + (m + n - 1)d] = 0$
673 Marks · March 2024 · Standardopen ↗
The ratio of the $10^{th}$ term to its $30^{th}$ 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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Let $a$ be the first term and $d$ be the common difference.
$\frac{a + 9d}{a + 29d} = \frac{1}{3}$
$\Rightarrow a=d \dots (i)$
$\frac{6}{2}(2a + 5d) = 42$
$\Rightarrow 2a + 5d = 14 \dots (ii)$
Solving (i) and (ii)
$a=2$ and $d=2$
683 Marks · March 2024 · Standardopen ↗
If the sum of the first $14$ terms of an A.P. is $1050$ and the first term is $10$, then find the $20^{\text{th}}$ term and the $n^{\text{th}}$ term.
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$\frac{14}{2}(20 + 13d) = 1050$
$\Rightarrow d = 10$
$\therefore a_{20} = 10 + 19 \times 10 = 200$
$a_n = 10 + (n - 1) 10 = 10n$
693 Marks · March 2024 · Standardopen ↗
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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$a = 5, a_n = 45, S_n = 400$
$\frac{n}{2}(5 + 45) = 400$
$\Rightarrow n = 16$
$5+15d=45$
$\Rightarrow d = \frac{40}{15} \text{ or } \frac{8}{3}$
5 Marks Questions
705 Marks · July 2023 · Standardopen ↗
Solve the equation for $x : - 4 + (-1) + 2 + \ldots + x = 437$.
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This is an Arithmetic Progression (AP).
$a = -4$, $d = -1 - (-4) = 3$.
Let $x$ be the $n^{th}$ term, so $a_n = x$.
Sum of $n$ terms $S_n = 437$.
$S_n = \frac{n}{2}[2a + (n-1)d]$
$437 = \frac{n}{2}[2(-4) + (n-1)3]$
$437 = \frac{n}{2}[-8 + 3n - 3]$
$437 = \frac{n}{2}[3n - 11]$
$874 = 3n^2 - 11n$
$3n^2 - 11n - 874 = 0$
Using quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $n = \frac{11 \pm \sqrt{(-11)^2 - 4(3)(-874)}}{2(3)}$
$n = \frac{11 \pm \sqrt{121 + 10488}}{6} = \frac{11 \pm \sqrt{10609}}{6} = \frac{11 \pm 103}{6}$
$n = \frac{11+103}{6} = \frac{114}{6} = 19$ or $n = \frac{11-103}{6} = \frac{-92}{6}$ (not possible as $n$ must be positive integer).
So, $n = 19$.
Now find $x = a_n = a_{19}$.
$a_n = a + (n-1)d$
$x = -4 + (19-1)3 = -4 + 18(3) = -4 + 54 = 50$.
So, $x = 50$.
715 Marks · July 2023 · Standardopen ↗
The sum of first $n$ terms of an AP is $5n^2 + 3n$. If its $n^{th}$ term is $168$, find $n$. Also, find the $20^{th}$ term of the AP.
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Given $S_n = 5n^2 + 3n$.
For $n=1$, $S_1 = 5(1)^2 + 3(1) = 5+3 = 8$. So, $a_1 = 8$.
For $n=2$, $S_2 = 5(2)^2 + 3(2) = 5(4) + 6 = 20+6 = 26$.
$S_2 = a_1 + a_2 \Rightarrow 26 = 8 + a_2 \Rightarrow a_2 = 18$.
Common difference $d = a_2 - a_1 = 18 - 8 = 10$.
Now, $a_n = 168$.
$a_n = a_1 + (n-1)d$
$168 = 8 + (n-1)10$
$160 = (n-1)10$
$16 = n-1 \Rightarrow n = 17$.
To find $20^{th}$ term, $a_{20} = a_1 + (20-1)d = 8 + 19(10) = 8 + 190 = 198$.
725 Marks · March 2023 · Standardopen ↗
The ratio of the $11^{\text{th}}$ term to $17^{\text{th}}$ term of an A.P. is $3:4$. Find the ratio of $5^{\text{th}}$ term to $21^{\text{st}}$ 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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Given $\frac{a + 10d}{a + 16d} = \frac{3}{4}$
$\Rightarrow 4a + 40d = 3a + 48d$
$\Rightarrow a = 8d$ (i)
therefore $\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d} = \frac{3}{7}$ using(i)
$a_5: a_{21} = 3:7$
$\frac{S_5}{S_{21}} = \frac{\frac{5}{2} (2a + 4d)}{\frac{21}{2} (2a + 20d)} = \frac{5 \times 20d}{21 \times 36d} = \frac{25}{189}$
Therefore, $S_5:S_{21}=25:189$
735 Marks · March 2023 · Standardopen ↗
Solve the equation for $x$: $1+4+7+ 10 + \dots + x = 287$
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$1+4+7+10 + \dots + x = 287$
$a = 1, d = 3$ Last term $= x$
$\Rightarrow 1 + (n - 1) 3 = x$
$3n - 2 = x \Rightarrow n = \frac{x + 2}{3}$
dots (i)
$S_n = 287$
$\frac{n}{2} [1 + x] = 287 \Rightarrow \frac{(\frac{x + 2}{3})}{2} [1 + x] = 287$
quad using (i)
$\frac{(x + 2)(x + 1)}{6} = 287$
$(x + 2)(x + 1) = 1722$
$x^2 + 3x + 2 = 1722$
$x^2 + 3x - 1720 = 0$
$(x + 43)(x - 40) = 0$
$\Rightarrow x = -43, 40$
$x \ne -43$ (as terms are increasing positive numbers)
$\therefore x = 40$
745 Marks · March 2023 · Standardopen ↗
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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$\frac{a + 10d}{a + 17d} = \frac{2}{3}$
$3a +30d = 2a + 34d \Rightarrow a = 4d$
Therefore, $\frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3}$
$\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a +20d]} = \frac{5[8d + 4d]}{21[8d + 20d]}$
$= \frac{5 \times 12d}{21 \times 28d} = \frac{5}{49}$ or $S_5: S_{21} = 5:49$
755 Marks · March 2023 · Standardopen ↗
Solve the equation : $-4+(-1)+2 + 5 + ..... + x = 437$.
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$-4+(-1)+2 + 5 + ..... + x = 437$
Here a = $-4$, d = $3$
$-4 + (n - 1)3 = x \Rightarrow n = \frac{x + 7}{3}$
$S_n = 437$
$\Rightarrow \frac{n}{2}(a+x) = 437$
$\Rightarrow \frac{(\frac{x+7}{3})}{2}(-4+x) = 437$
$\frac{(x+7)(x-4)}{6} = 437$
$x^2 + 3x - 28 = 437 \times 6 = 2622$
$x^2 + 3x - 2650 = 0$
$(x + 53)(x - 50) = 0$
$x \ne -53, x = 50$
765 Marks · March 2023 · Standardopen ↗
The sum of first seven terms of an A.P. is $182$. If its $4^{\text{th}}$ term and the $17^{\text{th}}$ term are in the ratio $1: 5$, find the A.P.
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Let $a$ be the first term and $d$ be the common difference.
$S_7 = 182 \Rightarrow \frac{7}{2}(2a + 6d) = 182$
$\Rightarrow 2a + 6d = \frac{182 \times 2}{7} = 52$
$a + 3d = 26$
quad ----- (i)
$\frac{a_4}{a_{17}} = \frac{1}{5} \Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}$
$\Rightarrow 5a + 15d = a+16d$
$4a = d$
quad --------(ii)
Solving (i) and (ii)
$a = 2$ and $d=8$
$\therefore \text{AP is } 2, 10, 18, 26, ....$
775 Marks · March 2023 · Standardopen ↗
OR
The sum of first $q$ terms of an A.P. is $63q - 3q^2$. If its $p^{\text{th}}$ term is $-60$, find the value of $p$. Also, find the $11^{\text{th}}$ term of this A.P.
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$S_q = 63q - 3q^2$
$\therefore S_1 = 60 \Rightarrow a_1 = 60$ ($1^{\text{st}}$ term)
$S_2 = 63(2) - 3(2)^2 = 126 - 12 = 114$
$a_1 + a_2 = 114 \Rightarrow a_2 = 114 - 60 = 54$
$d = a_2 - a_1 = 54 - 60 = -6$
$a_p = -60$
$60 + (p-1)d = -60$
$(p-1)(-6) = -120 \Rightarrow p=21$
$a_{11} = a + 10d = 60 + 10(-6) = 0$
785 Marks · March 2024 · Standardopen ↗
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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Sol. $a + a_8 = 32 \Rightarrow 2a + 7d = 32$ ----- (i)
$a \times a_8 = 60 \Rightarrow a(a + 7d) = 60$ ----- (ii)
Solving (i) & (ii), we get
$a=2$ or $a = 30$
and $d = 4$ or $d = -4$
First term and common difference of A.P. are $2$ and $4$ or $30$ and $-4$ respectively.
795 Marks · March 2024 · Standardopen ↗
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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Sol. Here $n = 40$,
$S_9 = \frac{9}{2} [2a +8d] = 153 \Rightarrow a + 4d = 17$ ----- (i)
and $S_{40} - S_{34} = 687$ or $a_{35} + a_{36} + a_{37} + a_{38} + a_{39} + a_{40} = 687$
$\Rightarrow 6a+219d = 687$ or $2a + 73d = 229$ ----- (ii)
solving (i) and (ii) to get $a = 5, d = 3$
Also, $S_{40} = \frac{40}{2} (10 + 39 \times 3) = 2540$
805 Marks · March 2025 · Standardopen ↗
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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Sol. Let first term $= a$ and common difference $= d$
ATQ, $(a + 2d) + (a + 6d) = 6$
$a + 4d = 3$
$a = 3-4d$
Also, $(a + 2d)(a + 6d) = 8$
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$9 - 4d^2 = 8$
$d = \pm \frac{1}{2}$
When $d = \frac{1}{2} \Rightarrow a = 1$
$S_{16} = \frac{16}{2}[2 \times 1 + 15 \times \frac{1}{2}]$
$= 76$
When $d = -\frac{1}{2} \Rightarrow a = 5$
$S_{16} = \frac{16}{2}[2 \times 5 + 15 \times (-\frac{1}{2})]$
$= 20$
815 Marks · March 2025 · Standardopen ↗
An AP consists of '$n$' terms whose $n^{\text{th}}$ 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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Let first term = $a$, common difference = $$\begin{aligned}& d = 2 \\ & a_n = a + (n - 1)2 = 4 \\ & text{ATQ, } a + 2n = 6 \\ & a = 6-2n \\ & text{ATQ, } S_n = \frac{n}{2}[2a + (n - 1)2] = -14 \\ & n[a + n - 1] = -14 \\ & n[6 - 2n + n - 1] = -14 \\ & n^2 - 5n - 14 = 0 \\ & Rightarrow n = 7 \\ & text{and } a = -8 \\ & S_{20} = \frac{20}{2}[2\times (-8) + 19 \times 2] \\ & = 220\end{aligned}$$
825 Marks · March 2025 · Standardopen ↗
The sum of the first six terms of an arithmetic progression is $42$. The ratio of the $10^{\text{th}}$ term to the $30^{\text{th}}$ term is $1 : 3$. Calculate the first and the thirteenth terms of the AP.
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Let first term = $a$ and common difference = $$\begin{aligned}& d \\ & text{ATQ, } \frac{a_{10}}{a_{30}} = \frac{a + 9d}{a + 29d} = \frac{1}{3} \\ & 3a + 27d = a + 29d \\ & 2a = 2d \Rightarrow a = d \\ & S_6 = \frac{6}{2}[2a + (6-1)d] = 42 \\ & 3[2a + 5a] = 42 \quad (\text{since } a=d) \\ & 3[7a] = 42 \\ & 21a = 42 \\ & a = 2 \\ & d = 2 \\ & a_{13} = a + 12d = 2 + 12 \times 2 = 2 + 24 = 26\end{aligned}$$

Word Problem & Applications

1 Mark Questions
831 Mark · March 2024 · Standardopen ↗
Three numbers in A.P. have the sum $30$. What is its middle term ?
  • (a)$4$
  • (b)$16$
  • (c)$10$
  • (d)$8$
Show SolutionHide Solution
(B) $10$
841 Mark · March 2025 · Standardopen ↗
Three numbers in AP have the sum $30$. What is its middle term ?
  • (a)$4$
  • (b)$10$
  • (c)$16$
  • (d)$8$
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(B) $10$
3 Marks Questions
853 Marks · July 2023 · Standardopen ↗
Find the sum of all integers between $50$ and $500$, which are divisible by $7$.
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$56, 63, ..., 497$
Here $a = 56$ and $d=7$
Let $a_n = 497$
$\Rightarrow 56 + (n-1) \times 7 = 497$
$\Rightarrow n = 64$
$S_{64} = \frac{64}{2} \times (56 + 497) = 17696$
863 Marks · July 2023 · Standardopen ↗
How many numbers lie between $10$ and $300$, which when divided by $4$ leave a remainder $3$ ? Also, find their sum.
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$11, 15, ..., 299$
Here $a = 11$ and $d=4$
Let $a_n = 299$
$\Rightarrow 11 + (n-1) \times 4 = 299$
$\Rightarrow n = 73$
$S_{73} = \frac{73}{2} \times (11 +299) = 11315$
873 Marks · March 2023 · Standardopen ↗
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 $30^{th}$ instalment ? What amount of loan has he paid after $30^{th}$ instalment ?
Show SolutionHide Solution
A.P formed is $1000, 1100, 1200, ...$
$a = 1000, d = 100$
$a_{30} = a + 29d = 3900$
Amount paid in $30^{th}$ instalment $= \text{\text{Rs} }3,900$
$S_{30} = \frac{30}{2}[2 \times 1000 + 29 \times 100] = 15 \times 4900 = 73,500$
$\therefore$ Total amount paid after $30^{th}$ instalment $= \text{\text{Rs} }73,500$
883 Marks · March 2024 · Standardopen ↗
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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Let the numbers be $a - d, a, a + d$
$\therefore a - d + a + a + d = 24$
$\Rightarrow a = 8$
Also, $(a - d)^2 + a^2 + (a + d)^2 = 194$
$\Rightarrow (8 - d)^2 + 8^2 + (8 + d)^2 = 194$
$\Rightarrow d^2 = 1 \Rightarrow d = \pm 1$
$\therefore$ Numbers are $7, 8, 9$ or $9,8,7$
893 Marks · March 2025 · Standardopen ↗
A sum of ₹2,000 is invested at $7\%$ per annum simple interest. Calculate the interests at the end of $1^{st}$, $2^{nd}$ and $3^{rd}$ year. Do these interests form an AP ? If so, find the interest at the end of the $27^{th}$ year.
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Interest at the end of $1^{st}$ year $= \frac{2000 \times 7 \times 1}{100} = \text{Rs}140$
Interest at the end of $2^{nd}$ year $= \frac{2000 \times 7 \times 2}{100} = \text{Rs}280$
Interest at the end of $3^{rd}$ year $= \frac{2000 \times 7 \times 3}{100} = \text{Rs}420$
$140, 280, 420, \dots$
Yes, Interests form an AP with first term = $140$ and common difference = $140$
Interest at the end of $27^{th}$ year $= 140 + 26 \times 140$
$= \text{Rs}3780$
903 Marks · March 2025 · Standardopen ↗
Find the sum of all $3$-digit natural numbers which are divisible by $11$.
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$3$ - digit natural numbers divisible by $11$ are
$110, 121, ..., 990$ ($1/2$)
Here first term $= 110$ and common difference $= 11$
$a_n = 990$
$\Rightarrow 110 + (n - 1) \times 11 = 990$ ($1$)
$\Rightarrow n = 81$ ($1/2$)
$S_{81} = \frac{81}{2} \times [110+990]$
$= 44550$ ($1$)
4 Marks Questions
914 Marks · March 2024 · Standardopen ↗
A school has decided to plant some endangered trees on $51^{st}$ World Environment Day in the nearest park. They have decided to plant those trees in few concentric circular rows such that each succeeding row has $20$ more trees than the previous one. The first circular row has $50$ trees.
Based on the above given information, answer the following questions :
(i) How many trees will be planted in the $10^{th}$ row?
(ii) How many more trees will be planted in the $8^{th}$ row than in the $5^{th}$ row?
(iii) (a) If $3200$ trees are to be planted in the park, then how many rows are required ?
OR
(b) If $3200$ trees are to be planted in the park, then how many trees are still left to be planted after the $11^{th}$ row?
figure for this question
Show SolutionHide Solution
Here $a = 50$ and $d = 20$
(i) Number of trees planted in $10^{th}$ row $= a_{10} = 50 + 9 \times 20 = 230$
(ii) $a_8 - a_5 = 3 \times 20 = 60$
(iii) (a) Let $S_n = 3200$
$\Rightarrow \frac{n}{2}[2 \times 50 + (n-1) \times 20] = 3200$
$\Rightarrow n^2 + 4n - 320 = 0$
$\Rightarrow (n + 20)(n - 16) = 0$
$n \neq -20$
$\therefore n = 16$
Hence, required number of rows are $16$
OR
(iii) (b) Required number of trees $= S_n - S_{11}$
$= 3200 - \frac{11}{2}[2 \times 50 + 10 \times 20]$
$= 3200 - 1550$
$= 1650$
Hence, number of trees left are $1650$
924 Marks · July 2024 · Standardopen ↗
Case Study – 1
Hari wants to participate in a $200$ m race. He can currently run that distance in $51$ seconds, and with each day of practice, he hopes to take $2$ seconds less than the previous day. He wants to do it in $31$ seconds.
Based on the above information, answer the following questions:
(i) Write the A.P. which represents the above situation.
(ii) Find the minimum number of days he needs to practice to achieve the goal.
(iii) (a) Find the expression for the $n^{th}$ term of the A.P.
OR
(b) If he wants to do it in $21$ seconds, how many minimum days will he take ?
Show SolutionHide Solution
Sol. (i) $51, 49, 47, 45, \dots, 31$
(ii) Here $a = 51 \& d = -2$
$31 = 51 + (n - 1) (-2)$
$\Rightarrow n = 11$
So, minimum $11$ days he need to practice to achieve the goal.
(iii) (a) $a_n = 51 + (n - 1) (-2)$
$a_n = 53 - 2n$
OR
(b) $21 = 51 + (n - 1) (-2)$
$n = 16$
So, minimum $16$ days he need to practice to achieve the goal.
934 Marks · March 2024 · Standardopen ↗
Case Study - 2
Treasure Hunt is an exciting and adventurous game where participants follow a series of clues/numbers/maps to discover hidden treasures. Players engage in a thrilling quest, solving puzzles and riddles to unveil the location of the coveted prize.
While playing a treasure hunt game, some clues (numbers) are hidden in various spots collectively forming an A.P. If the number on the $n^{th}$ spot is $20+4n$, then answer the following questions to help the players in spotting the clues :
(i) Which number is on first spot?
(ii) (a) Which spot is numbered as $112$?
OR
(b) What is the sum of all the numbers on the first $10$ spots?
(iii) Which number is on the $(n - 2)^{th}$ spot ?
figure for this question
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(i) Number on the first spot $$\begin{aligned}& = 20 + 4 \times 1 = 24 \\ & \text{(ii) (a) } 20 + 4n = 112 \\ & \Rightarrow n = 23 \\ & \text{OR} \\ & \text{(ii) (b) } d = 4 \\ & S_{10} = \frac{10}{2} [2 \times 24 + 9 \times 4] \\ & = 420 \\ & \text{(iii) Number on the } (n - 2)^{th} \text{ spot } = 20 + 4(n - 2) \\ & = 12 + 4n\end{aligned}$$
944 Marks · March 2024 · Standardopen ↗
Case Study – 3
A road roller is a compactor-type engineering vehicle, used to compact soil, gravel, concrete, etc, in the construction of roads and foundations. They are also used at landfills or in agriculture. A company started making road rollers $10$ years ago and increased its production uniformly by a fixed number every year. The company produces $800$ rollers in the $6^{th}$ year and $1130$ rollers in the $9^{th}$ year.
Based on the above information, answer the following questions :
(i) What is the company's production in the first year ?
(ii) What was the increase in the company's production every year?
(iii) (a) What was the company's production in the $8^{th}$ year?
OR
(b) What was the company's total production in the first $6$ years?
Show SolutionHide Solution
(i) $$\begin{aligned}& (a + 8d) - (a + 5d) = 330 \Rightarrow d = 110 \\ & a+5\times110 = 800 \Rightarrow a = 250 \\ & (ii) d=110 \\ & (iii) (a)\end{aligned}$$a_8 = 250 + 7
times 110
= 1020
OR
(iii) (b) $$\begin{aligned}& S_6 = \frac{6}{2} [2 \times 250 + 5 \times 110] \\ & = 3150\end{aligned}$$
954 Marks · July 2025 · Standardopen ↗
Case Study - 2
In the month of September, villagers of Ankurhut were falling ill with high temperature. Paracetamol was one of the highest sold medicines during that phase. A survey was conducted to estimate the overall sale of Paracetamol of each pharmacy during the last $7$ days. It was observed that the number of Paracetamol sold in different shops were all 3-digit numbers, divisible by $13$, taken in order.
Based on the information given above, answer the following questions :
(i) How many Paracetamols were sold by the $7^{th}$ pharmacy?
(ii) What was the difference between the number of Paracetamols sold by the $14^{th}$ and the $9^{th}$ pharmacy?
(iii) (a) How many Paracetamols were sold by the $9^{th}$ pharmacy from the last?
OR
(iii) (b) What was the total number of Paracetamols sold in that week?
figure for this question
Show SolutionHide Solution
(i) A.P. formed is
$104, 117, 130, \dots$ with $a = 104$ and $d = 13$
$a_7 = 104 + 6 \times 13 = 182$
(ii) $a_{14} - a_9 = 5 \times 13 = 65$
(iii) (a) Last term of A.P. is $988$ and $d = -13$
$a_9 = 988 + 8 \times (-13) = 884$
OR
(b) Last term of A.P. is $988$ and $d = -13$
$988 = 104 + (n - 1) \times 13$
$\Rightarrow n = 69$
$S_{69} = \frac{69}{2} \times (104+988)$
$= 37674$
964 Marks · March 2025 · Standardopen ↗
Case Study - 1: A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being $300$ metres. The organizers decide to increase the distance of each subsequent round by $50$ metres. The total number of rounds planned is $10$. (i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. (ii) Determine the distance of the $8^{th}$ round. (iii) (a) Find the total distance run after completing all $10$ rounds. OR (iii) (b) If a runner completes only the first $6$ rounds, what is the total distance run by the runner?
figure for this question
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A.P: $300, 350, 400 \dots$ (i) $a_4 = 450, a_5 = 500, a_6 = 550$. (ii) $a_8 = 300 + 7 \times 50 = 650$ m. (iii)(a) $S_{10} = \frac{10}{2}(2 \times 300 + 9 \times 50) = 5250$ m. (iii)(b) $S_6 = \frac{6}{2}(2 \times 300 + 5 \times 50) = 2250$ m.
974 Marks · March 2025 · Standardopen ↗
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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Sol. The ages of the participants form the following AP
$8, 8\frac{1}{3}, 8\frac{2}{3}, 9, ...$
where first term $= 8$ and common difference $= \frac{1}{3}$
Let the number of participants be $n$
$S_n = \frac{n}{2}[2 \times 8 + (n-1)\frac{1}{3}] = 168$
$n^2 + 47n - 1008 = 0$
$\Rightarrow n = 16$
$\therefore$ the age of the eldest participant $= 8 + 15 \times \frac{1}{3} = 13$ years
984 Marks · March 2025 · Standardopen ↗
Cable cars at hill stations are one of the major tourist attractions. On a hill station, the length of cable car ride from base point to top most point on the hill is $5000$ m. Poles are installed at equal intervals on the way to provide support to the cables on which car moves.
The distance of first pole from base point is $200$ m and subsequent poles are installed at equal interval of $150$ m. Further, the distance of last pole from the top is $300$ m.
Based on above information, answer the following questions using Arithmetic Progression :
(i) Find the distance of $10^{th}$ pole from the base.
(ii) Find the distance between $15^{th}$ pole and $25^{th}$ pole.
(iii) (a) Find the time taken by cable car to reach $15^{th}$ pole from the top if it is moving at the speed of $5$m/sec and coming from top.
OR
(iii) (b) Find the total number of poles installed along the entire journey.
figure for this question
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AP formed is $200, 350, 500, ...$
(i) Distance of $10^{th}$ pole from base = $A_{10}$
$= 200+9\times 150$
$= 1550$ m
(ii) Distance between $15^{th}$ pole and $25^{th}$ pole = $a_{25} - A_{15}$
$= 10 \times 150 = 1500$ m
(iii) (a) Distance of $15^{th}$ pole from the top = $300 + 14 \times 150$
$= 2400$ m
Time taken by cable car = $\frac{2400}{5} = 480$ seconds or $8$ minutes
OR
(iii) (b) Distance of last pole from the base = $(5000 – 300)$ m = $4700$ m
$\therefore a_n = 4700$
$\Rightarrow 200+ (n – 1)150 = 4700$
Solving, we get $n = 31$
994 Marks · March 2025 · Standardopen ↗
In an equilateral triangle of side 10 cm, equilateral triangles of side 1 cm are formed as shown in the figure below, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row and so on. Based on given information, answer the following questions using Arithmetic Progression. (i) How many triangles will be there in bottom most row? (ii) How many triangles will be there in fourth row from the bottom? (iii) (a) Find the total number of triangles of side 1 cm each till $8^{th}$ row. OR (iii) (b) How many more number of triangles are there from $5^{th}$ row to $10^{th}$ row than in first 4 rows? Show working.
figure for this question
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Given A.P. is 1, 3, 5, ...
(i) $a_{10} = 1 + 9 \times 2 = 19$
(ii) $a_4$ (from bottom) $= 19 + 3 \times (-2) = 13$
(iii) (a) $S_8 = \frac{8}{2} \times [2 \times 1 + 7 \times 2] = 64$
(iii) (b) Number of triangles from $5^{th}$ row to $10^{th}$ row $= S_{10} - S_4 = \frac{10}{2} \times [2 \times 1 + 9 \times 2] - \frac{4}{2} \times [2 \times 1 + 3 \times 2] = 84$. Number of triangles in first 4 rows, $S_4 = \frac{4}{2} \times [2 \times 1 + 3 \times 2] = 16$. Required number of triangles $= 84 - 16 = 68$
1004 Marks · March 2025 · Standardopen ↗
In order to organise, Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below: The length of innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane. Based on given information, answer the following questions, using concept of Arithmetic Progression. (i) What is the length of the 6th lane? (ii) How long is the 8th lane than that of 4th lane? (iii) (a) While practicing for a race, a student took one round each in first six lanes. Find the total distance covered by the student. OR (iii) (b) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student.
figure for this question
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Here AP is 400, 407.6, 415.2, ... (i) $a_6 = 400 + 5(7.6) = 438$ m (1 mark). (ii) $a_8 - a_4 = 30.4$ m (1 mark). (iii) $S_6 = \frac{6}{2}(2 \times 400 + 5 \times 7.6) = 2514$ m (1 + 1 marks). OR (iii) Total distance covered = $S_8 - S_3 = \frac{8}{2}(2 \times 400 + 7 \times 7.6) - \frac{3}{2}(2 \times 400 + 2 \times 7.6) = 2190$ m (1 + 1 marks).
1014 Marks · March 2025 · Standardopen ↗
To inculcate the good habit of savings in her children, Reema brought a piggy bank and after putting a ₹ $10$ coin in it, she handed it over to her daughter Amisha and asked as to put money in it from her pocket money at the beginning of every week. Amisha put two ten rupee coins at the beginning of next (second) week and in this way increases her savings by one ₹ $10$ coin every week. Based on the above, answer the following questions : (a) How many coins were added in the piggy bank at the beginning of $5^{th}$ week ? (b) How many ₹ $10$ coins will be there in the piggy bank after the end of $7$ weeks ? (c) (i) If the piggy bank can hold a maximum of $300$ ₹ $10$ coins, after how many weeks it would be full ? OR (c) (ii) Find the total amount of money in the piggy bank at the end of $20$ weeks.
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(a) $a_5 = 1 + 4(1) = 5$
(b) $S_7 = \frac{7}{2} [2 + 6 \times 1] = 28$ coins
(c) (i) $S_n = \frac{n}{2} [2 + (n - 1) \times 1] = 300 \implies n = 24$ weeks
OR
(c) (ii) $S_{20} = 10[20 + 19 \times 10] = \text{\text{Rs} } 2100$
5 Marks Questions
1025 Marks · March 2023 · Standardopen ↗
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?
figure for this question
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Here $a = 22, d = – 1$ $S_n = 250$
$\therefore 250 = \frac{n}{2} [44 + (n – 1) (-1)]$
$\Rightarrow n^2 - 45n + 500 = 0$
$\Rightarrow (n-25) (n – 20) = 0$
$n\neq 25 \therefore n = 20$
logs in top row = $a_{20} = 22 + 19 (- 1) = 3$
1035 Marks · March 2023 · Standardopen ↗
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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ATQ,
$32+36 +40 + ..... = 2,000$
$a = 32, d =4$ ($\frac{1}{2} + \frac{1}{2}$)
Let 'n' be the number of months.
$S_n = 2000$
$\frac{n}{2}[2(32) + (n - 1) 4] = 2000$ (1)
$n^2 + 15n - 1000 = 0$ (1)
$n^2 + 40n - 25n - 1000 = 0$
$(n + 40) (n - 25) = 0$ (1)
$\Rightarrow n = -40, n = 25$
(rejecting)
$\therefore n = 25$ (1)
So, Prerna will save ₹2,000, in $25$ months.
1045 Marks · March 2023 · Standardopen ↗
Find the sum of integers between $100$ and $200$ which are (i) divisible by $9$ (ii) not divisible by $9$.
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(i) Integers divisible by $9$ are $108, 117, 126, ....., 198$
a = $108$, d = $9$
a + $(n-1)d = 198$
$\Rightarrow 108 + (n - 1)9 = 198 \Rightarrow n = 11$
$S_{11} = \frac{n}{2}(a+l) = \frac{11}{2}(108 + 198)$
$= 1683$
(ii) Sum of all integers = $\frac{99}{2}(101 + 199) = \frac{99}{2} \times 300 = 14850$
Sum of integers not divisible by $9 = 14850 - 1683 = 13167$
1055 Marks · March 2024 · Standardopen ↗
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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Let his starting salary is $a$ and annual increment be $d$.
A.T.Q
$a+3d=15000$ ---------(i)
$a+9d=18000$ ----------(ii)
On solving equations (i) & (ii)
$a = 13500$ & $d = 500$
Starting salary = ₹13,500
Annual increment = ₹500

Two APs

1 Mark Questions
1061 Mark · July 2024 · Standardopen ↗
Two A.P.s have the same first term. The common difference of the first A.P. is $- 3$ and of the second A.P. is $- 5$. The difference of the $6^{th}$ term of the second A.P. from that of the first A.P. is :
  • (a)$2$
  • (b)$-8$
  • (c)$-10$
  • (d)$10$
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(D) $10$

General

1 Mark Questions
1071 Mark · July 2025 · Standardopen ↗
If $\cot A = \frac{7}{12}$, then the value of $(\cos A + \sin A) \cosec A$ is:
  • (a)$\frac{5}{12}$
  • (b)$\frac{19}{7}$
  • (c)$\frac{19}{12}$
  • (d)$\frac{49}{144}$
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(B) $\frac{19}{12}$