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If the $p^{\text {th }}$ term of an A.P. be $\frac{1}{q}$ and $q^{\text {th }}$ term be $p q^{t h}$ terms will be
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Verified Answer
The correct answer is:
$\frac{p q+1}{2}$
Since $T_p=a+(p-1) d=\frac{1}{q}$\ldots(i)
and $T_q=a+(q-1) d=\frac{1}{p}$\ldots(ii)
From (i) and (ii), we get
$a=\frac{1}{p q}$ and $d=\frac{1}{p q}$
Now sum of $p q$ terms
$=\frac{p q}{2}\left[\frac{2}{p q}+(p q-1) \frac{1}{p q}\right]$
$=\frac{p q}{2} \cdot \frac{2}{p q}\left[1+\frac{1}{2}(p q-1)\right]$ $=\left[\frac{2+p q-1}{2}\right]=\frac{p q+1}{2}$
Note : Students should remember this question as a formula.
and $T_q=a+(q-1) d=\frac{1}{p}$\ldots(ii)
From (i) and (ii), we get
$a=\frac{1}{p q}$ and $d=\frac{1}{p q}$
Now sum of $p q$ terms
$=\frac{p q}{2}\left[\frac{2}{p q}+(p q-1) \frac{1}{p q}\right]$
$=\frac{p q}{2} \cdot \frac{2}{p q}\left[1+\frac{1}{2}(p q-1)\right]$ $=\left[\frac{2+p q-1}{2}\right]=\frac{p q+1}{2}$
Note : Students should remember this question as a formula.
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