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

Step $1:$ Finding common difference.
Given, $a_p=\frac{1}{q}$ and $a_q=\frac{1}{p}$
We know, $n^{th}$ term of A.P
$a_n=a+(n-1)d$
$a_p=a+(p-1)d=\frac{1}{q}\cdots \left(i\right)$
$a_q=a+(q-1)d=\frac{1}{p}\cdots \left(ii\right)$
Now, substracting equation $\left(i\right)$ and $\left(ii\right)$ we get
$(p-1)d-(q-1)d=\frac{1}{q}-\frac{1}{p}$
$\Rightarrow [p-1-q+1]d=\frac{p-q}{pq}$
$\Rightarrow (p-q)d=\frac{p-q}{pq}$
$\therefore d=\frac{1}{pq}$

Step $2:$ Finding first term.
Substituting the value of $d$ in equation $\left(i\right)$ we get
$a+(p-1)\frac{1}{pq}=\frac{1}{q}$
$\Rightarrow a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}$
$\therefore a=\frac{1}{pq}$

Step $3:$ Finding sum of first $pq$ term of an $A.P.$
As we know, sum of $n$ tems in $A.P$
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{pq}=\frac{pq}{2}[\frac{2}{pq}+(pq-1\left)\frac{1}{pq}\right]$
$\Rightarrow S_{pq}=\frac{1}{2}[2+pq-1]$
$\therefore S_{pq}=\frac{1}{2}[pq+1]$

Final answer: Hence, the sum of first $pq$ terms of the $A.P.$ is $\frac{1}{2}(pq+1)$.