Search any question & find its solution
Question:
Answered & Verified by Expert
In an A.P., if $p^{\text {th }}$ term is $\frac{1}{q}$ and $q^{\text {th }}$ term is $\frac{1}{p}$, prove that the sum of first $p q$ terms is $\frac{1}{2}(p q+1)$, where $p \neq q$
Solution:
2098 Upvotes
Verified Answer
Let a be the first term and $d$ be the common difference of an A.P. then
$T_p=\frac{1}{q} \Rightarrow a+(p-1) d=\frac{1}{q} \quad \ldots \text { (i) }$
$T_q=\frac{1}{p} \Rightarrow a+(q-1) d=\frac{1}{p} \quad \ldots \text { (ii) }$
Eqn (i) - eqn (ii) gives,
$(p-q) d=\frac{1}{q}-\frac{1}{p}=\frac{p-q}{p q} \Rightarrow d=\frac{1}{p q}$
Putting $d=\frac{1}{p q}$ in (i), we get
$\begin{aligned}
&a+(p-1) \frac{1}{p q}=\frac{1}{q} \Rightarrow a+\frac{1}{q}-\frac{1}{p q}=\frac{1}{q} \\
&\Rightarrow a=\frac{1}{p q}
\end{aligned}$
Now, $S_{p q}=\frac{p q}{2}[2 a+(p q-1) d]$
$=\frac{p q}{2}\left[\frac{2}{p q}+(p q-1) \frac{1}{p q}\right]$
$=\frac{p q}{2}\left[\frac{2}{p q}+1-\frac{1}{p q}\right]=\frac{p q}{2}\left(\frac{1}{p q}+1\right)$
$S_{p q}=\frac{1}{2}(p q+1)$
$T_p=\frac{1}{q} \Rightarrow a+(p-1) d=\frac{1}{q} \quad \ldots \text { (i) }$
$T_q=\frac{1}{p} \Rightarrow a+(q-1) d=\frac{1}{p} \quad \ldots \text { (ii) }$
Eqn (i) - eqn (ii) gives,
$(p-q) d=\frac{1}{q}-\frac{1}{p}=\frac{p-q}{p q} \Rightarrow d=\frac{1}{p q}$
Putting $d=\frac{1}{p q}$ in (i), we get
$\begin{aligned}
&a+(p-1) \frac{1}{p q}=\frac{1}{q} \Rightarrow a+\frac{1}{q}-\frac{1}{p q}=\frac{1}{q} \\
&\Rightarrow a=\frac{1}{p q}
\end{aligned}$
Now, $S_{p q}=\frac{p q}{2}[2 a+(p q-1) d]$
$=\frac{p q}{2}\left[\frac{2}{p q}+(p q-1) \frac{1}{p q}\right]$
$=\frac{p q}{2}\left[\frac{2}{p q}+1-\frac{1}{p q}\right]=\frac{p q}{2}\left(\frac{1}{p q}+1\right)$
$S_{p q}=\frac{1}{2}(p q+1)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.