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If $(p+q)^{\text {th }}$ term of a G.P. be $m$ and $(p-q)^{t h}$ term be $n$, then the $p^{\text {th }}$ term will be
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The correct answer is:
$\sqrt{m n}$
Given that $m=a r^{p+q-1}$ and $n=a r^{p-q-1}$
$r^{p+q-1-p+q+1}=\frac{m}{n} \Rightarrow$ $r=\left(\frac{m}{n}\right)^{1 /(2 q)}$
and $a=\frac{m}{\left(\frac{m}{n}\right)^{(p+q-1) / 2 q}}$
$=a r^{\rho-1}=\frac{m}{\left(\frac{m}{n}\right)^{(p+q-1) / 2 q}}\left(\frac{m}{n}\right)^{(p-1) / 2 q}$
Now $p^{\text {th }}$ term $=m\left(\frac{m}{n}\right)^{(p-1) / 2 q-(p+q-1) /(2 q)}$ $=m\left(\frac{m}{n}\right)^{-1 / 2}=m^{1-1 / 2} n^{1 / 2}$
Aliter : As we know each term in a G.P. is geometric mean of the terms equidistant from it. Here $(p+q)^{\text {th }}$ and
$(p-q)^{\text {th }}$ terms are equidistant from $p^{\text {th }}$ term i.e. at a distance of $q$. Therefore, $p^{\text {th }}$ term will be G.M. of $(p+q)^{\text {th }}$ and $(p-q)^{t h}$ i.e. $\sqrt{m n}$
$r^{p+q-1-p+q+1}=\frac{m}{n} \Rightarrow$ $r=\left(\frac{m}{n}\right)^{1 /(2 q)}$
and $a=\frac{m}{\left(\frac{m}{n}\right)^{(p+q-1) / 2 q}}$
$=a r^{\rho-1}=\frac{m}{\left(\frac{m}{n}\right)^{(p+q-1) / 2 q}}\left(\frac{m}{n}\right)^{(p-1) / 2 q}$
Now $p^{\text {th }}$ term $=m\left(\frac{m}{n}\right)^{(p-1) / 2 q-(p+q-1) /(2 q)}$ $=m\left(\frac{m}{n}\right)^{-1 / 2}=m^{1-1 / 2} n^{1 / 2}$
Aliter : As we know each term in a G.P. is geometric mean of the terms equidistant from it. Here $(p+q)^{\text {th }}$ and
$(p-q)^{\text {th }}$ terms are equidistant from $p^{\text {th }}$ term i.e. at a distance of $q$. Therefore, $p^{\text {th }}$ term will be G.M. of $(p+q)^{\text {th }}$ and $(p-q)^{t h}$ i.e. $\sqrt{m n}$
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