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If the $(2 \mathrm{p})^{\text {th }}$ term of a $\mathrm{H}$. $\mathrm{P}$. is $\mathrm{q}$ and the $(2 \mathrm{q})^{\text {th }}$ term is $\mathrm{p},$ then the $2(\mathrm{p}+\mathrm{q})^{\mathrm{th}}$ term is-
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Verified Answer
The correct answer is:
$\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}$
If a is the first term and d is the common difference of the associated A.P. $\frac{1}{\mathrm{q}}=\frac{1}{\mathrm{a}}+(2 \mathrm{p}-1) \mathrm{d}, \frac{1}{\mathrm{p}}=\frac{1}{\mathrm{a}}+(2 \mathrm{q}-1) \mathrm{d}$
$\Rightarrow \mathrm{d}=\frac{1}{2 \mathrm{pq}}$
If h is the $2(\mathrm{p}+\mathrm{q})^{\mathrm{th}}$ term $\frac{1}{\mathrm{~h}}=\frac{1}{\mathrm{a}}+(2 \mathrm{p}+2 \mathrm{q}-1) \mathrm{d}$
$=\frac{1}{\mathrm{q}}+\frac{1}{\mathrm{p}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}$
$\Rightarrow \mathrm{d}=\frac{1}{2 \mathrm{pq}}$
If h is the $2(\mathrm{p}+\mathrm{q})^{\mathrm{th}}$ term $\frac{1}{\mathrm{~h}}=\frac{1}{\mathrm{a}}+(2 \mathrm{p}+2 \mathrm{q}-1) \mathrm{d}$
$=\frac{1}{\mathrm{q}}+\frac{1}{\mathrm{p}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}$
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