JEE Main 2014 — Sequences and Series Question with Solution

Given that $200<S_9<220$

As we know sum of $n$ terms of A.P whose first term is  $a$ and common difference $d$ is $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
$\Rightarrow 200<\frac{9}{2}\left(2a+8d\right)<220$

$\Rightarrow \frac{200}{9}<a+4d<\frac{220}{9}$

As we know $n^{th}$ term of A.P is $T_n=a+\left(n-1\right)d$

Also   $T_2=a+d=12$

$\Rightarrow 4d=48-4a$

$\therefore \frac{200}{9}<a+48-4a<\frac{220}{9}$

$\Rightarrow \frac{200}{9}-48<-3a<\frac{220}{9}-48$

$\Rightarrow \frac{-232}{9}<-3a<-{\frac{212}{9}}_{{}_{}}$

OR   $\frac{212}{27}<a<\frac{232}{27}$

i.e. $7\frac{23}{27}<a<8\frac{16}{27}$

$\Rightarrow a=8$

$\Rightarrow d=4$

$T_4=a+3d=8+12=20$