JEE Main 2014 — Sequences and Series Question with Solution

We know that the $n^{th}$ term and sum of $n$terms of an $A.P.$ with first term $a$ and common difference $d$ are respectively $a+\left(n-1\right)d$ and $\frac{n}{2}\left[2a+\left(n-1\right)d\right].$

Let, the number of terms in the given $A.P.$ be $2n$ then there are $\frac{2n}{2}=n$ even terms and $n$ odd terms.

Then, $T_1=a$ and $T_{2n}=a+(2n-1)d$

Given $T_{2n}-T_1=\frac{21}{2}$

$\Rightarrow \left(2n-1\right)d=\frac{21}{2}$

$\Rightarrow 2nd-d=\frac{21}{2} ...\left(i\right)$

Also, the sum of odd terms i.e. $a+\left(a+2d\right)+\left(a+4d\right)+...$ is

$\frac{n}{2}\left[2a+\left(n-1\right)2d\right]=24$

$\Rightarrow 2a+\left(n-1\right)2d=\frac{48}{n} ...\left(ii\right)$

And, the sum of even terms i.e. $\left(a+d\right)+\left(a+3d\right)+\left(a+5d\right)+...$ is

$\frac{n}{2}\left[2\left(a+d\right)+\left(n-1\right)2d\right]=30$

$\text{⇒}\frac{n}{2}\left[2a+\left(n-1\right)2d+2d\right]=30$

Put the value from equation $\left(ii\right)$ to get

$\frac{n}{2}\left[\frac{48}{n}+2d\right]=30$

$\Rightarrow 24+dn=30$

$\Rightarrow nd=6 ...\left(iii\right)$

Put this value in the equation $\left(i\right),$ to get

$2\times 6-d=\frac{21}{2}$

$\Rightarrow d=\frac{24-21}{2}=\frac{3}{2}$

Now, put $d$ in the equation $\left(iii\right),$ to get

$\Rightarrow n\times \frac{3}{2}=6$

$\Rightarrow n=4$

$\Rightarrow 2n=8$

Thus, the number of terms in the given $A.P.$ is $8.$