**Question:**

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:**

It is known that the general term of an A.P. is *a**n* = *a* + (*n* – 1)*d*

∴ According to the given information,

$p^{\text {th }}$ term $=a_{p}=a+(p-1) d=\frac{1}{q}$ $\ldots(1)$

$q^{\text {th }}$ term $=a_{q}=a+(q-1) d=\frac{1}{p}$ $\ldots(2)$

Subtracting (2) from (1), we obtain

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

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

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

$\Rightarrow d=\frac{1}{p q}$

Putting the value of *d* in (1), we obtain

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

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

$\therefore S_{p q}=\frac{p q}{2}[2 a+(p q-1) d]$

$=\frac{p q}{2}\left[\frac{2}{p q}+(p q-1) \frac{1}{p q}\right]$

$=1+\frac{1}{2}(p q-1)$

$=\frac{1}{2} p q+1-\frac{1}{2}=\frac{1}{2} p q+\frac{1}{2}$

$=\frac{1}{2}(p q+1)$

Thus, the sum of first $p q$ terms of the A.P. is $\frac{1}{2}(p q+1)$.

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