How many terms of the AP 3,7,11,15,... willl make the sum 406 ?
(a) 10 (b) 12 (c) 14 (d) 20

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Solution

$A P : 3, 7, 11, 15, . . a = 3 d = 7 - 3 = 4 S_{n} = 406 W e k n o w t h a t, S_{n} = \frac{n}{2} [2 a + (n - 1) d] \frac{n}{2} [2 (3) + (n - 1) 4] = 406 \frac{n}{2} [6 + 4 n - 4] = 406 \frac{n}{2} [4 n + 2] = 406 n [2 n + 1] = 406 2 n^{2} + n - 406 = 0 2 n^{2} - 28 n + 29 n - 406 = 0 2 n (n - 14) + 29 (n - 14) = 0 (2 n + 29) (n - 14) = 0 n = \frac{- 29}{2} o r 14$

n cannot be negative.

So n = 14

Hence, 14 terms will make the sum 406.

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