If the $p^{th}$ term of an A.P. is $\frac{1}{q}$ and  $q^{th}$ term of an A.P. is $\frac{1}{p}$. Prove that the sum of first  $pq$ term of the A.P. is $\left[\frac{pq+1}{2}\right]$. 

We need to prove that sum of first 𝑝𝑞 term of the A.P. is given that $\left[\frac{pq+1}{2}\right]$  given that $p^{th}$ term of an A.P. is $\frac{1}{q}$ and  $q^{th}$ term of an A.P. is  $\frac{1}{p}$.

Let the first term and the common difference between of the A.P. be 𝑎 and 𝑑 respectively. It is known that the 𝑛𝑡ℎ term of an AP is given by

$a_n=a+(n-1)d$, where  $a_n$ is  ${n }^{th}$the term, $a$ is the first term, $n$is the term to be found and $d$ is the common difference of the A.P. 

Therefore, the  $p^{th}$ term  $\left(a_p\right)$ is given by 

$\begin{array}{l}{a}_p=a+(p-1)d=\frac{1}{q}\\ \Rightarrow a+(p-1)d=\frac{1}{q} .......\left(i\right)\end{array}$

$\text{ Similarly, the }{q}^{\text{th }}\text{ term term }\left(a_q\right)\text{ is given by }$

$\begin{array}{l}{a}_q=a+(q-1)d=\frac{1}{p}\\ \Rightarrow a+(q-1)d=\frac{1}{p} ....\left(ii\right)\end{array}$

Now, on solving the two equations, we get

$\begin{array}{l}a=\frac{1}{pq}\text{ and }\\ d=\frac{1}{pq}\end{array}$

Now, it is known that the sum of an AP is given by

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\text{,} \mathrm{where} S_n\text{ is the sum of terms, }n\text{ is the number of terms, }a\text{ is the }$first term and 𝑑 is the common difference of AP. On substituting 𝑛 as 𝑝𝑞, we get the sum as

$\begin{array}{l}\Rightarrow S_{pq}=\frac{pq}{2}\left[2\left(\frac{1}{pq}\right)+(pq-1)\left(\frac{1}{pq}\right)\right]\\ \Rightarrow S_{pq}=\frac{1}{2}[2+(pq-1\left)\right]\\ \Rightarrow S_{pq}=\frac{1}{2}[pq+1]\end{array}$

Hence, proved.