If the p^{th} term of an A.P. is frac{1}{q} a | vedclass

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(C) Let the first term be $a$ and the common difference be $d$.Given that the $p^{th}$ term $T_p = a + (p - 1)d = \frac{1}{q} \dots (i)$And the $q^{th

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$\frac{pq + 1}{2}$

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(C) Let the first term be $a$ and the common difference be $d$.Given that the $p^{th}$ term $T_p = a + (p - 1)d = \frac{1}{q} \dots (i)$And the $q^{th}$ term $T_q = a + (q - 1)d = \frac{1}{p} \dots (ii)$Subtracting equation $(ii)$ from $(i)$:$(p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$Thus,$d = \frac{1}{pq}$.Substituting $d$ in equation $(i)$:$a + (p - 1)\frac{1}{pq} = \frac{1}{q} \implies a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q} \implies a = \frac{1}{pq}$.The sum of $pq$ terms $S_{pq} = \frac{pq}{2} [2a + (pq - 1)d]$.Substituting the values of $a$ and $d$:$S_{pq} = \frac{pq}{2} [2(\frac{1}{pq}) + (pq - 1)(\frac{1}{pq})]$$S_{pq} = \frac{pq}{2} [\frac{2 + pq - 1}{pq}] = \frac{pq}{2} [\frac{pq + 1}{pq}] = \frac{pq + 1}{2}$.

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

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