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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)

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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)

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Ex 5.3, 9 You are here

Ex 5.3, 10 (i)

Ex 5.3, 10 (ii) Important

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Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

Last updated at March 28, 2023 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 = π/2 (2π+(πβ1)π) Sum of first 7 terms = 49 S7 = 7/2 (2π+(πβ1)π) 49 = 7/2 (2π+(7β1)π) 49 = 7/2 (2π+6π) (49 Γ 2)/7 "= 2a + 6d" "14 = 2a + 6d" (14 β 6π)/2=π a = 7 β 3d Sum of first 17 terms = 289 S17 = 17/2 (2π+(17β1)π) 289 = 17/2 (2a + (17 β 1) d) 289 = 17/2 (2a + 16d) (289 Γ 2)/17 = 2a + 16d 34 = 2a + 16 d (34 β 16π)/2 = a a = 17 β 8d From (1) and (2) 7 β 3d = 17 β 8d 8d β 3d = 17 β 7 5d = 10 d = 10/5 d = 2 Putting value of d in (1) a = 7 β 3d a = 7 β 3 Γ2 a = 7 β 6 a = 1 Hence, a = 1 & d = 2 We need to find sum of first n terms We can use formula Sn = π/π (2a + (n β 1) d) Putting a = 1 & d = 2 = π/2 (2 Γ 1+(πβ1)2) = π/2(2+2πβ2) = π/2 (0 + 2n) = π/2 Γ 2n = n2