If $(p + q)^{th}$ term and $(p - q)^{th}$ term of a G.P. be $m$ and $n,$ then the $p^{th}$ term will be ___.

$\frac{m}{n}$

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$\sqrt{m n}$

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Solution

The correct option is B
$\sqrt{m n}$

Given that
$a_{p + q} = a r^{p + q - 1} = m$ and $a_{p - q} = a r^{p - q - 1} = n .$

$\Rightarrow m \times n = a r^{p + q - 1}$ x $a r^{p - q - 1}$

$= a^{2} r^{2 (p - 1)}$

$= (a r^{p - 1})^{2}$

$\Rightarrow \sqrt{m n} = a r^{p - 1} = a_{p}$

Thus, the $p^{th}$ term of the GP is $\sqrt{m n} .$

Aliter: Each term in a G.P. is the geometric mean of the terms equidistant from it. Here, $(p + q)^{t h}$ and $(p - q)^{t h}$ terms are equidistant from the $p^{t h}$ term.
$∴$ $p^{t h}$ term ( $\sqrt{m n}$ ) will be G.M. of $(p + q)^{t h}$ and $(p - q)^{t h}$ terms.

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