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

Since, In $A.P$,

If '$a$' be the first term and '$d$' be the common difference then $n$-th term,

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

Given,

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

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

$\frac{1}{q}-\frac{1}{p}=[a+(p-1\left)d\right]-[a+(q-1\left)d\right]$

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

$\Rightarrow \frac{p-q}{pq}=dp-dq$

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

$\therefore d=\frac{1}{pq}$ ---(3)

From (1)

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

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

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

$\Rightarrow a=\frac{1}{pq}$ ---(4)

Now, the sum of $n$-terms in $A.P$ is, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

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

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

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

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

$\therefore S_{pq}=\frac{1}{2}[pq+1]$

Hence, proved.