Ex 5.3, 9 - Chapter 5 Class 10 Arithmetic Progressions
Last updated at Dec. 13, 2024 by Teachoo
Last updated at Dec. 13, 2024 by Teachoo
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 = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Sum of first 7 terms = 49 S7 = 7/2 (2𝑎+(𝑛−1)𝑑) 49 = 7/2 (2𝑎+(7−1)𝑑) 49 = 7/2 (2𝑎+6𝑑) (49 × 2)/7 "= 2a + 6d" "14 = 2a + 6d" (14 − 6𝑑)/2=𝑎 a = 7 – 3d Sum of first 17 terms = 289 S17 = 17/2 (2𝑎+(17−1)𝑑) 289 = 17/2 (2a + (17 – 1) d) 289 = 17/2 (2a + 16d) (289 × 2)/17 = 2a + 16d 34 = 2a + 16 d (34 − 16𝑑)/2 = a a = 17 – 8d From (1) and (2) 7 – 3d = 17 – 8d 8d – 3d = 17 – 7 5d = 10 d = 10/5 d = 2 Putting value of d in (1) a = 7 – 3d a = 7 – 3 ×2 a = 7 – 6 a = 1 Hence, a = 1 & d = 2 We need to find sum of first n terms We can use formula Sn = 𝒏/𝟐 (2a + (n – 1) d) Putting a = 1 & d = 2 = 𝑛/2 (2 × 1+(𝑛−1)2) = 𝑛/2(2+2𝑛−2) = 𝑛/2 (0 + 2n) = 𝑛/2 × 2n = n2
Ex 5.3
Ex 5.3, 1 (ii)
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Ex 5.3, 9 You are here
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo