If 64, 27, 36 are the P^{text{th}}, Q^{text{t | vedclass

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(C) Let $a$ be the first term and $r$ be the common ratio of a $GP$. The $P^{	ext{th}}$, $Q^{	ext{th}}$, and $R^{	ext{th}}$ terms are $ar^{P-1}$, $

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$3R$

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(C) Let $a$ be the first term and $r$ be the common ratio of a $GP$. The $P^{	ext{th}}$, $Q^{	ext{th}}$, and $R^{	ext{th}}$ terms are $ar^{P-1}$, $ar^{Q-1}$, and $ar^{R-1}$ respectively.According to the question:$ar^{P-1} = 64$ $(i)$$ar^{Q-1} = 27$ (ii)$ar^{R-1} = 36$ (iii)Dividing $(i)$ by (ii):$r^{P-Q} = \frac{64}{27} = \left(\frac{4}{3}ight)^3$ (iv)Dividing (ii) by (iii):$r^{Q-R} = \frac{27}{36} = \frac{3}{4}$Raising to the power of $3$:$r^{3(Q-R)} = \left(\frac{3}{4}ight)^3 = \left(\frac{4}{3}ight)^{-3}$ $(v)$Multiplying (iv) and $(v)$:$r^{P-Q} 	imes r^{3Q-3R} = \left(\frac{4}{3}ight)^3 	imes \left(\frac{4}{3}ight)^{-3} = 1$$r^{P+2Q-3R} = r^0$Therefore, $P+2Q-3R = 0$, which implies $P+2Q = 3R$.

If $64, 27, 36$ are the $P^{\text{th}}$, $Q^{\text{th}}$, and $R^{\text{th}}$ terms of a $GP$, then $P+2Q$ is equal to

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