Let a be the first term and \(r\) be the common ration of a GP. \(\therefore P^{\text {th }}, Q^{\text {th }}\) and \(R^{\text {th }}\) terms of a GP are respectively ar \({ }^{\mathrm{P}-1}\), ar \(\mathrm{Q}^{-1}\) and \({a r}^{\mathrm{R}-1}\).
According to question, \(a r^{P-1}=64 \ldots . .\) (i)
\(a r^{Q-1}=27 \ldots . .\) (ii)
\(a r^{R-1}=36 \ldots . .\) (iii)
Dividing Eq. (i) by Eq. (ii), we get
\(\mathrm{r}^{\mathrm{P}-\mathrm{Q}}=\left(\frac{4}{3}\right)^{3} \ldots \ldots\) (iv)
Dividing Eq. (ii) by Eq. (iii), we get
\(\begin{array}{l}
\mathrm{r}^{\mathrm{Q}-\mathrm{R}}=\frac{3}{4} \\
\Rightarrow \mathrm{r}^{3 \mathrm{Q}-3 \mathrm{R}}=\left(\frac{3}{4}\right)^{3}
\end{array}\)
Multiplying Eq. (iv) and Eq. (v), we get
\(\begin{array}{l}
\mathrm{r}^{\mathrm{P}-\mathrm{Q}} \times \mathrm{r}^{3 \mathrm{Q}-3 \mathrm{R}}=1 \\
\Rightarrow \mathrm{r}^{\mathrm{P}-\mathrm{Q}+3 \mathrm{Q}-3 \mathrm{R}}=1 \\
\Rightarrow \mathrm{r}^{\mathrm{P}+2 \mathrm{Q}-3 \mathrm{R}}=\mathrm{r}^{0} \\
\Rightarrow \mathrm{P}+2 \mathrm{Q}-3 \mathrm{R}=0 \\
\Rightarrow \mathrm{P}+2 \mathrm{Q}=3 \mathrm{R}
\end{array}\)