# ICSE Class 10 Arithmetic Progression (A.P.) PYQs with Solution

Arithmetic Progression, commonly known as A.P., is a fundamental concept in mathematics. It plays a significant role in various mathematical and real-world applications. In ICSE Class 10, understanding and solving A.P. problems are essential for scoring well in mathematics. In this article, we will explore a collection of ICSE Class 10 Arithmetic Progression (A.P.) PYQs with solution.

ICSE Class 10 Arithmetic Progression (A.P.) PYQs (previous years’ questions) with solution provide valuable practice material for students preparing for their exams. By solving these questions, you will develop a strong foundation in A.P. concepts, improve your problem-solving skills, and gain confidence in tackling similar questions. Remember to practice regularly and seek additional resources to enhance your understanding. With dedication and consistent effort, you can excel in ICSE Class 10 Mathematics and achieve your desired results.

## ICSE Class 10 Arithmetic Progression (A.P.) PYQs with Solution

1. Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60… is 300? Hence find the sum of all the terms of the Arithmetic Progression. 

Solution: 20th term , 3150

Step-by-step Explanation:

Let 300 be nth term.

a = 15, d = 30 – 15 = 15

We know, Tn =a + (n – 1) d

Therefore, 300 = 15 + (n – 1) 15

300 = 15 + 15n – 15

300 = 15n

15n = 300

n = 20

Hence, 300 is the 20th term.

$S_n=\frac n2(a+l)\\S_{20}=\frac{20}2(15+300)\\S_{20}=10\times315\\S_{20}=3150$

2. The 5th term and the 9th term of an Arithmetic Progression are 4 and -12 respectively.
Find:
(a) the first term
(b) common difference
(c) sum of 16 terms of the AP. 

Solution: (a) 20 (b) -4 (c) -160

Step-by-step Explanation:

$Given,\;T_5=4\;and\;T_9=-12\\So,\;a+4d\;=\;4\;……(1)\\a+8d\;=\;-12\;…….(2)\\subtracting\;equation\;(2)\;from\;(1),\;we\;have\\4d-8d=4+12\\-4d=16\\d=-4\\putiing\;the\;value\;of\;d\;in\;(1)\\a+4(-4)=4\\a=4+16\\a=20\\(a)\;first\;term\;=\;20\\(b)\;common\;difference=-4\\(c)\;S_{16}=\frac n2\lbrack2a+(n-1)d\rbrack\\\Rightarrow\frac{16}2\lbrack2\times20+15\times(-4)\rbrack\\\Rightarrow8(40-60)\\\Rightarrow-160$

Watch ICSE Class 10 Arithmetic Progression (A.P.) PYQs Video Solution Here.

3. The nth term of an Arithmetic Progression (A.P.) is 2n + 5. The 10th term is:

(a) 7
(b) 15
(c) 25
(d) 45 

Solution: (c)

Step-by-step Explanation:

$Given,\;nth\;term\;of\;an\;A.P.=2n+5\\T_n=2n+5\\T_{10}=2\times10+5\\T_{10}=25$

4. The sum of the first 20 terms of the Arithmetic Progression 2, 4, 6, 8,….is :
(a) 400
(b) 840
(c) 420
(d) 800 [2021 Semester-1]

Solution: (c)

Step-by-step Explanation:

$S_n=\frac n2\lbrack2a+(n-1)d\rbrack\\\therefore S_{20}=\frac{20}2\lbrack2\times2+(20-1)2\rbrack\\S_{20}=10\;\lbrack4+38\rbrack\\S_{20}=420$

5. The first four terms of an Arithmetic Progression (A. P.), whose first term is 4 and common difference is -6, are:
(a) 4, -10, -16, -22
(b) 4, 10, 16, 22
(c) 4, -2, -8, -14
(d) 4, 2, 8, 14 [2021 Semester-1]

Solution: (c)

Step-by-step Explanation:

$Given,\\First\;term\;a\;=\;4\\common\;difference\;d\;=\;-6\\\therefore T_2=4-6=-2\\T_3=-2-6=-8\\T_4=-8-6=-14\\$

6. In an Arithmetic Progression (A.P.) if, first term is 5, common difference is – 3 and then nth term is – 7, then n is equal to:
(a) 5
(b) 17
(c) – 13
(d) 7 [2021 Semester-1]

Solution: (a)

Step-by-step Explanation:

$Given,\\first\;term\;a=\;5,\;\\common\;difference\;d=\;-\;3\;\\nth\;term\;=\;-\;7,\\\therefore T_n=a+(n-1)d\\-7=5+(n-1)(-3)\\-7=5+-3n+3\\3n=15\\n=5\\$

7. If 70, 75, 80, 85 are the first four terms of an Arithmetic Progression. Then the 10th term is:
(a) 35
(b) 25
(c) 115
(d) 105 [2021 Semester-1]

Solution: (c)

Step-by-step Explanation:

$first\;term\;a=\;70,\;\\common\;difference\;d=\;a_2-a_1=75-70=5\\T_n=a+(n-1)d\\\therefore T_{10}=70+(10-1)5\\=70+45\\=115\\\\$

8. If the 6th term of an A.P. is equal to four times its first term and the sum of first six terms is 75, find the first term and the common difference. 

Solution: a = 5 , d = 3

Step-by-step Explanation:

$Given,\\T_6=4\times a\\\Rightarrow a+5d=4a\\\Rightarrow5d=3a\\\Rightarrow d=\frac{3a}5\\S_6=75\\\Rightarrow\frac n2\lbrack2a+(n-1)d\rbrack=75\\\Rightarrow\frac62\lbrack2a+5\times\frac{3a}5\rbrack=75\\\Rightarrow3\lbrack2a+3a\rbrack=75\\\Rightarrow15a=75\\\Rightarrow a=5\\Hence,\;d=\frac{3\times5}5=3\\\\$

9. In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find
(i) first term
(ii) common difference
(iii) sum of the first 20 terms. 

Solution: (i) -1, (ii) 3 , (iii) 550

Step-by-step Explanation:

$Given,\\T_4=8\\T_6=14\\so,\;a+3d=8\;…..(1)\\a+5d=14\;……\;(2)\\subtracting\;(2)\;from\;(1)\;we\;have,\\-2d=-6\\d=3\\putting\;the\;value\;of\;d\;in\;(1)\\a+3\times3=8\\a=-1\\Therefore,\\(i)\;a=-1\\(ii)\;d=3\\(iii)\;S_{20}=\frac{20}2\lbrack2\times(-1)+(20-1)(3)\rbrack\\\Rightarrow S_{20}=10\;\lbrack-2+57\rbrack\\\Rightarrow S_{20}=550\\\\\\$

10. The sum of the first three terms of an Arithmetic Progression (A:P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference.  

Solution: first term= 14, common difference= 12

Step-by-step Explanation:

$Let\;the\;first\;three\;terms\;of\;an\;A.P.\;be\;(a-d),\;a,\;(a+d)\\Given,\;\\a-d+a+a+d=42\\\Rightarrow3a=42\\\Rightarrow a=14\\and\;(a-d)(a+d)=52\\\Rightarrow a^2-d^2=52\\\Rightarrow14^2-d^2=52\\\Rightarrow196-52=d^2\\\Rightarrow144=d^2\\\Rightarrow d=12\\\\\\$

ICSE Class 10 Maths PYQs with Solution

11. If (k – 3), (2k + 1) and (4k + 3) are three consecutive terms of an A.P., find the value of k.  

Solution: k=2

Step-by-step Explanation:

$(k\;–\;3),\;(2k\;+\;1)\;and\;(4k\;+\;3)\;are\;three\;consecutive\;terms\;of\;an\;A.P.\\Therefore,\;\\a_3-a_2=a_2-a_1\\\Rightarrow(4k+3)-(2k+1)=(2k+1)-(k-3)\\\Rightarrow4k+3-2k-1=2k+1-k+3\\\Rightarrow2k+2=k+4\\\Rightarrow k=2$

12. The 4th term of an A. P. is 22 and 15th term is 66. Find the first term and the common difference. Hence find the sum of the series to 8 terms. 

Solution: a = 10, d = 4, S8 =192

Step-by-step Explanation:

$Given,\;\\T_4=22\;\\T_{15}=66\;\\a+3d=22\;……(1)\\a+14d=66\;……(2)\\subtracting\;(1)\;from\;(2)\;we\;have,\\11d=44\\d=4\\putting\;the\;value\;of\;d\;in\;(1)\\a+3\times4=22\\a=10\\S_8=\frac82\lbrack2\times10+(8-1)(4)\rbrack\\\Rightarrow S_8=4\lbrack20+28\rbrack\\\Rightarrow S_8=192$

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