How many terms of the AP 3, 7, 11, 15, ...

Question:

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

 

Solution:
(c) 14
Here, = 3 and d = (7-3) = 4
Let the sum of terms be 406 .
Then,
 we have:
$S n=\frac{n}{2}[2 a+(n-1) d]=406$
$\Rightarrow \quad \frac{n}{2}[2 \times 3+(n-1) \times 4]=406$
$\Rightarrow n[3+2 n-2]=406$
$\quad \Rightarrow 2 n^{2}+n-406=0$
$\Rightarrow 2 n^{2}-28 n+29 n-406=0$
$\Rightarrow 2 n(n-14)+29(n-14)=0$
$\Rightarrow(2 n+29)(n-14)=0$
$\Rightarrow n=14 \quad(\because n$ can 't be a fraction $)$
Hence, 14 terms will make the sum 406.

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