Question

If $64,27,36$ are the Pth Qth and $R$ th terms of a GP, then $P+2Q$ is equal to

• A. R
• B. 2R
• C. 3R
• D. 4R

Answer 1

Answer: (C)

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