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Ex 5.3, 9 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at March 28, 2023 by Teachoo
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
Ex 5.3, 16 Important
Ex 5.3, 17
Ex 5.3, 18 Important
Ex 5.3, 19 Important
Ex 5.3, 20 Important
Chapter 5 Class 10 Arithmetic Progressions
Serial order wise
Transcript
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
