Last updated at March 22, 2021 by Teachoo

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Ex 5.3, 9 If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. We know that Sn = ๐/2 (2๐+(๐โ1)๐) Sum of first 7 terms = 49 S7 = 7/2 (2๐+(๐โ1)๐) 49 = 7/2 (2๐+(7โ1)๐) 49 = 7/2 (2๐+6๐) (49 ร 2)/7 "= 2a + 6d" "14 = 2a + 6d" (14 โ 6๐)/2=๐ a = 7 โ 3d Sum of first 17 terms = 289 S17 = 17/2 (2๐+(17โ1)๐) 289 = 17/2 (2a + (17 โ 1) d) 289 = 17/2 (2a + 16d) (289 ร 2)/17 = 2a + 16d 34 = 2a + 16 d (34 โ 16๐)/2 = a a = 17 โ 8d From (1) and (2) 7 โ 3d = 17 โ 8d 8d โ 3d = 17 โ 7 5d = 10 d = 10/5 d = 2 Putting value of d in (1) a = 7 โ 3d a = 7 โ 3 ร2 a = 7 โ 6 a = 1 Hence, a = 1 & d = 2 We need to find sum of first n terms We can use formula Sn = ๐/๐ (2a + (n โ 1) d) Putting a = 1 & d = 2 = ๐/2 (2 ร 1+(๐โ1)2) = ๐/2(2+2๐โ2) = ๐/2 (0 + 2n) = ๐/2 ร 2n = n2

Ex 5.3

Ex 5.3, 1 (i)

Ex 5.3, 1 (ii)

Ex 5.3, 1 (iii) Important

Ex 5.3, 1 (iv)

Ex 5.3, 2 (i)

Ex 5.3, 2 (ii)

Ex 5.3, 2 (iii) Important

Ex 5.3, 3 (i)

Ex 5.3, 3 (ii)

Ex 5.3, 3 (iii)

Ex 5.3, 3 (iv) Important

Ex 5.3, 3 (v)

Ex 5.3, 3 (vi) Important

Ex 5.3, 3 (vii)

Ex 5.3, 3 (viii) Important

Ex 5.3, 3 (ix)

Ex 5.3, 3 (x)

Ex 5.3, 4

Ex 5.3, 5

Ex 5.3, 6 Important

Ex 5.3, 7

Ex 5.3, 8

Ex 5.3, 9 You are here

Ex 5.3, 10 (i)

Ex 5.3, 10 (ii) Important

Ex 5.3, 11 Important

Ex 5.3, 12

Ex 5.3, 13

Ex 5.3, 14 Important

Ex 5.3, 15 Deleted for CBSE Board 2022 Exams

Ex 5.3, 16 Important Deleted for CBSE Board 2022 Exams

Ex 5.3, 17 Deleted for CBSE Board 2022 Exams

Ex 5.3, 18 Important Deleted for CBSE Board 2022 Exams

Ex 5.3, 19 Important Deleted for CBSE Board 2022 Exams

Ex 5.3, 20 Important Deleted for CBSE Board 2022 Exams

Chapter 5 Class 10 Arithmetic Progressions (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.