Question

If the $p^{th}$ term of an A.P be $\frac{1}{q}$ and $q^{th}$ term be $\frac{1}{p}$, then sum of its $p{q}^{th}$ term will be

• A. $\frac{\left(pq-1\right)}{2}$
• B. $\frac{\left(1-pq\right)}{2}$
• C. $\frac{\left(pq+1\right)}{2}$
• D. $-\frac{\left(pq+1\right)}{2}$

Answer 1

The correct option is C

$\frac{\left(pq+1\right)}{2}$

Explanation of the correct option.

Step $1$: Find the value of first term and common difference.

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

We know that $a_n=a+(n-1)d$

Given: $a_p=\frac{1}{q}$ and $a_q=\frac{1}{p}$

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

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

Equation $\left(1\right)-\left(2\right)$,

$$\begin{gathered} \Rightarrow \frac{1}{q}-\frac{1}{p}=pd-d-qd+d \\ \Rightarrow \frac{\left(p-q\right)}{pq}=d\left(p-q\right) \\ \Rightarrow d=\frac{1}{pq} \end{gathered}$$

Substitute the value of $d$ in Equation $\left(1\right)$,

,$$\begin{gathered} \frac{1}{q}=a+(p-1)\frac{1}{pq} \\ \Rightarrow \frac{1}{q}=a+\frac{1}{q}-\frac{1}{pq} \\ \Rightarrow a=\frac{1}{pq} \end{gathered}$$

Step $2$: Find the required sum.

We know that sum of $n^{th}$ term of an A.P. is given by,

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

$$\begin{gathered} \Rightarrow S_{pq}=\frac{pq}{2}\left(2\left(\frac{1}{pq}\right)+(pq-1)\left(\frac{1}{pq}\right)\right) \\ \Rightarrow S_{pq}=\frac{pq}{2}\left(\frac{2+pq-1}{pq}\right) \\ \Rightarrow S_{pq}=\frac{\left(1+pq\right)}{2} \end{gathered}$$

Therefore, the sum of $p{q}^{th}$ is $\frac{1+pq}{2}$.

Hence option$\left(C\right)$ is the correct option.