The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP. [CBSE 2009]

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Solution

$Sum of 9 term s of AP = 81 Sum of n terms of AP is S_{n} = \frac{n}{2} [2 a + (n - 1) d] 81 = \frac{9}{2} [2 a + 8 d] 2 a + 8 d = 18 . . . . . (1) Sum of 1 st 20 terms is 400 400 = \frac{20}{2} [2 a + 19 d] 2 a + 19 d = 40 . . . . . (2) Subtract (1) from (2) d = 2 and a = 1 Therefore, first term is 1 and common difference is 2 .$

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The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP. [CBSE 2009]

Sum of 9 terms of AP=81Sum of n terms of AP is Sn=n22a+n-1d81=922a+8d2a+8d=18 .....1Sum of 1st 20 terms is 400400=2022a+19d2a+19d=40 .....2Subtract 1 from 2d=2 and a=1Therefore, first term is 1 and common difference is 2.