Question

In an A.P. if $p^{th}$ term is $\frac{1}{q}\text{and}{q}^{th}$term is

$\frac{1}{p}$. Prove that the sum of first pq terms is

$\frac{1}{2}$ $(pq+1),$ where p $\ne$ q

Answer 1

Let a be the first term and d be the common difference of given A.P.

Here $a_p=\frac{1}{q}and{a}_q=\frac{1}{p}$

$\therefore a+(p-1)d=\frac{1}{q}anda+(q-1)d=\frac{1}{p}$

$a+pd-d=\frac{1}{q}$ ......(i)

$anda+qd-d=\frac{1}{p}$ ......(ii)

$a+pd-d(a+qd-d)=\frac{1}{q}-\frac{1}{p}$

$\Rightarrow a+pd-d-a-qd+d=\frac{p-q}{pq}$

$\Rightarrow (p-q)d=\frac{p-q}{pq}$

$\Rightarrow d=\frac{p-q}{pq}\times \frac{1}{p-q}$

= $\frac{1}{pq}$

Putting value of d in (i)

$a+(p-1)\times \frac{1}{pq}=\frac{1}{q}$

$\Rightarrow a=\frac{1}{q}-\frac{p-1}{pq}=\frac{p-p+1}{pq}=\frac{1}{pq}$

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

$\therefore s_{pq}=\frac{pq}{2}[2\times \frac{1}{pq}+(pq-1)\times \frac{1}{pq}]$

= $\frac{pq}{2}[\frac{2}{pq}+\frac{pq-1}{pq}]=\frac{pq}{2}\left[\frac{2+pq-1}{pq}\right]$

= $\frac{pq}{2}\left[\frac{pq+1}{pq}\right]=\frac{pq+1}{2}$

$\therefore s_{pq}=\frac{1}{2}pq+1.$