Question

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

General term of an A.P. is given by

$a_n=a+(n-1)d$

Sum of n terms of an A.P. is given by

$S_n=\frac{n}{2}2a+(n-1)d$

where $a$ is the first term and $d$ is the common difference.

Given $a_p=\frac{1}{q}$

$\Rightarrow a+(p-1)d=\frac{1}{q}$ ...(1)

$a_q=\frac{1}{p}$

$\Rightarrow a+(q-1)d=\frac{1}{p}$ ...(2)

Subtracting eq(2) from eq (1), we get

$(p-1)d-(q-1)d=\frac{1}{q}-\frac{1}{p}$

$\Rightarrow (p-1-q+1)d=\frac{p-q}{pq}$

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

$\Rightarrow d=\frac{1}{pq}$

Putting the value of $d$ in (1), we get
$a+(p-1)\frac{1}{pq}=\frac{1}{q}$

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

Sum of $pq$ terms of an A.P. is

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

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

$=1+\frac{1}{2}(pq-1)$

$=\frac{1}{2}pq+1-\frac{1}{2}$

$=\frac{1}{2}pq+\frac{1}{2}$

$=\frac{1}{2}(pq+1)$

So, the sum of $pq$ terms of A.P. is $\frac{1}{2}(pq+1)$