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(a) Find the sum of first 30 terms of AP :- 30, -24, -18, .... (b) In an AP if Sn = n (4n + 1), then find the AP. |
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Answer» (a) a1 = -30, a2 = -24 d = a2 - a1 = -24 - (-30) = 6 \(\therefore\) sum of first 30 terms is S30 = 30/2 [2a + (30 - 1)] = 15 [2 x -30 + 29 x 6] = 15 [-60 + 174] = 15 x 114 = 1710 (b) Given sum of n terms of A.P. is Sn = n(4n + 1) T1 = S1 = 1(4 + 1) = 5 T2 = S2 - S1 = 2(4 x 2 + 1) - 5 = 18 - 5 = 13 T3 = S3 - S2 = 3(4 x 3 + 1) - 18 = 39 -18 = 21 T4 = S4 - S3 = 4(4 x 4 + 1) - 39 = 68 - 39 = 29 Hence, required AP is 5, 13, 21, 29..... |
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