Question

If $a,b,c$ are distinct and are in $G.P.$ with common ratio $r$ and $a,2b,3c$ are in $A.P.$, then $r$ equals

• A. $1$
• B. $\frac{1}{3}$
• C. $-\frac{1}{3}$
• D. $-1$

Answer 1

Answer: (B)

$\frac{1}{3}$

Given $a,b,c$ are in GP
$\Rightarrow b=ar,c=a{r}^2$
Also, $\because a,2b,3c$ are in $A.P.$
$\Rightarrow 4b=a+3c$
$\Rightarrow 4ar=a+3a{r}^2$
$\Rightarrow 4r=1+3r^2$

$\Rightarrow 3r^2-4r+1=0$

$\Rightarrow 3r^2-3r-r+1=0$

$\Rightarrow 3r(r-1)-(r-1)=0$

$\Rightarrow (3r-1)(r-1)=0$

$\Rightarrow r=1$ or $\frac{1}{3}$

But, $r\ne 1$ as it is a common ratio

$\therefore r=\frac{1}{3}$