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The number of terms of the A.P. 3, 7, 11, 15, ……. to be taken so that the sum is 406 is

A. 5 B. 10 C. 12 D. 14 Answer: Option D
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Solution(By Apex Team)

The A.P. is 3, 7, 11, 15, …… Where a = 3, d = 7 – 3 = 4 and sum Sn = 406 $\begin{aligned}&\therefore S_n=\frac{n}{2}[2a+(n-1)d]\\ &\Rightarrow406=\frac{n}{2}[2\times3\times(n-1)\times4]\\ &\Rightarrow812=n(6+4n-4)\\ &\Rightarrow812=n(4n+2)\\ &\Rightarrow4n^2+2n-812=0\\ &\Rightarrow2n^2+n-406=0\\ &\Rightarrow2n^2+29n-28n-406=0\\ &\Rightarrow n(2n+29)-14(2n+29)=0\\ &\Rightarrow(2n+29)(n-14)=0\\ &\therefore n=14\text{ or }\frac{-29}{2}\\ &\text{But}\ n=\frac{-29}{2}\ \text{is not possible}\end{aligned}$

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