How many terms of the AP 3, 7, 11, 15, ...... will make the sum 406 ?

Foundation

Mathematics Foundation

Sum of n terms of an AP

Question

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Let S_n = 406. Then, $\frac{n}{2} [2 \times 3 + (n - 1) \times 4] = 406$

$\Rightarrow \frac{n}{2} (6 + 4 n - 4) = 406 \Rightarrow \frac{n}{2} (4 n + 2) = 406$

$∴ n = \frac{- 1 \pm \sqrt{1 + 3248}}{4} = \frac{- 1 + \sqrt{3249}}{4}$ [neglecting negative value]

$\Rightarrow n = \frac{- 1 + 57}{4} = \frac{56}{4} = 14 .$