If the pth and qth terms of a GP are q and p, respectively, then (p-q)th term is

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${(\frac{p^{q}}{q^{p}})}^{\frac{1}{p - q}}$

${(\frac{q^{p}}{p^{q}})}^{\frac{1}{p - q}}$

${(\frac{q^{q}}{p^{p}})}^{\frac{1}{p - q}}$

None of these

answer is A.

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Detailed Solution

let fist term be A and common ratio be R. Given, pth term, T_p = q. and q th term, T_q = p .

$\begin{array}{l} {AR}^{p - 1} = q and {AR}^{q - 1} = p . . . . . . (i) \\ ∴ \frac{{AR}^{p - 1}}{{AR}^{q - 1}} = \frac{q}{p} \end{array}$

$\Rightarrow R^{p - q} = \frac{q}{p} \Rightarrow R = {(\frac{q}{p})}^{\frac{1}{p - q}}$

On putting the value of -R in Eq. (i), we get

$A \cdot {(\frac{q}{p})}^{\frac{p - 1}{p - q}} = q \Rightarrow A = q \cdot {(\frac{p}{q})}^{\frac{p - 1}{p - q}}$

Now, (p+g)th term, $T_{p + q} = {AR}^{p + q - 1}$

$\begin{array}{l} = q \cdot {(\frac{p}{q})}^{\frac{p - 1}{p - q}} \times {(\frac{q}{p})}^{\frac{p + q - 1}{p - q}} \\ = \frac{q^{1 - \frac{p - 1}{p - q} + \frac{p + q - 1}{p - q}}}{\frac{p + q - 1}{p - q} - \frac{p - 1}{p - q}} = \frac{q^{\frac{p - q - p + 1 + p + q - 1}{p - q}}}{p^{\frac{p + q - 1 - p + 1}{p - q}}} \\ = \frac{q^{\frac{p}{p - q}}}{\frac{q}{p^{p - q}}} = {(\frac{q^{p}}{p^{q}})}^{\frac{1}{p - q}} = {(\frac{q^{p}}{p^{q}})}^{\frac{1}{(p - q)}} \end{array}$