###
**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**

## Show Answer

Solution(By Apex Team)

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

## Related Questions On Progressions

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### Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

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### The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is

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