If the (p + q)^{th} term of a geometric progr | vedclass

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(B) Let the first term be $a$ and the common ratio be $r$.Given that the $(p + q)^{th}$ term is $m = a r^{p + q - 1}$ and the $(p - q)^{th}$ term is $

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$\sqrt{mn}$

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(B) Let the first term be $a$ and the common ratio be $r$.Given that the $(p + q)^{th}$ term is $m = a r^{p + q - 1}$ and the $(p - q)^{th}$ term is $n = a r^{p - q - 1}$.Dividing the two equations: $\frac{m}{n} = \frac{a r^{p + q - 1}}{a r^{p - q - 1}} = r^{(p + q - 1) - (p - q - 1)} = r^{2q}$.Thus,$r^{2q} = \frac{m}{n}$,which implies $r = (\frac{m}{n})^{1/(2q)}$.The $p^{th}$ term is $T_p = a r^{p - 1}$.Note that $m 	imes n = (a r^{p + q - 1}) 	imes (a r^{p - q - 1}) = a^2 r^{2p - 2} = (a r^{p - 1})^2$.Therefore,$(T_p)^2 = mn$,which gives $T_p = \sqrt{mn}$.

If the $(p + q)^{th}$ term of a geometric progression is $m$ and the $(p - q)^{th}$ term is $n$,then what is the $p^{th}$ term?

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