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Ex 5.3, 9 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)

Last updated at March 22, 2021 by

Ex 5.3, 9 - If sum of first 7 terms of an AP is 49, and 17 term is 289

Ex 5.3, 9 - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.3, 9 - Chapter 5 Class 10 Arithmetic Progressions - Part 3

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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 = 𝑛/2 (2π‘Ž+(π‘›βˆ’1)𝑑) Sum of first 7 terms = 49 S7 = 7/2 (2π‘Ž+(π‘›βˆ’1)𝑑) 49 = 7/2 (2π‘Ž+(7βˆ’1)𝑑) 49 = 7/2 (2π‘Ž+6𝑑) (49 Γ— 2)/7 "= 2a + 6d" "14 = 2a + 6d" (14 βˆ’ 6𝑑)/2=π‘Ž a = 7 – 3d Sum of first 17 terms = 289 S17 = 17/2 (2π‘Ž+(17βˆ’1)𝑑) 289 = 17/2 (2a + (17 – 1) d) 289 = 17/2 (2a + 16d) (289 Γ— 2)/17 = 2a + 16d 34 = 2a + 16 d (34 βˆ’ 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 Γ—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 Γ— 1+(π‘›βˆ’1)2) = 𝑛/2(2+2π‘›βˆ’2) = 𝑛/2 (0 + 2n) = 𝑛/2 Γ— 2n = n2

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