Question

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

• A. $2R$
• B. $3R$
• C. $4R$
• D. $R$

Answer 1

Answer: (B)

Let $a$ be the first term and $r$ be the common ration of a GP.
$\therefore$ $P^{th},Q^{th}$ 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
$r^{P-Q}={\left(\frac{4}{3}\right)}^3$ ..... (iv)

Dividing Eq. (ii) by Eq. (iii), we get
$r^{Q-R}=\frac{3}{4}$
$\Rightarrow r^{3Q-3R}={\left(\frac{3}{4}\right)}^3$

Multiplying Eq. (iv) and Eq. (v), we get
$r^{P-Q}\times r^{3Q-3R}=1$
$\Rightarrow r^{P-Q+3Q-3R}=1$
$\Rightarrow r^{P+2Q-3R}=r^0$
$\Rightarrow P+2Q-3R=0$
$\Rightarrow P+2Q=3R$