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

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14 Explanation: Let Sn = 406. Then, `n/2 [2 xx 3 + (n - 1) xx 4] = 406 $\Rightarrow$ n/2 (6 + 4n - 4) = 406 $\Rightarrow$ n/2 (4n + 2) = 406` $\Rightarrow$ n(2n + 1) = 406 $\Rightarrow$ 2n2 + n – 406 = 0 $\therefore$ n = (-1 +- sqrt(1 + 3248))/4 = (-1 + sqrt(3249))/4 ...[Neglecting negative value] $\Rightarrow$ n = (-1 + 57)/4 = $\frac{56}{4}$ = 14

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