Question

If the first three terms of the sequence $\frac{1}{16},a,b,\frac{1}{6}$ are in $G.P.$ and the last three terms are in $H.P.$ where $(a>0,b>0)$, then the value of $a+b$ is

• A. $8$
• B. $\frac{7}{36}$
• C. $\frac{5}{36}$
• D. $5$

Answer 1

The correct option is B $\frac{7}{36}$

$\frac{1}{16},a,b$ are in $G.P.$
$\Rightarrow 16a^2=b\cdots \left(1\right)$
$a,b,\frac{1}{6}$ are in $H.P.$
$\Rightarrow \frac{2a}{6a+1}=b\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$16a^2=\frac{2a}{6a+1}$
$\Rightarrow 48a^2+8a-1=0$
$\Rightarrow a=-\frac{1}{4},\frac{1}{12}$
$\therefore a=\frac{1}{12}$ and $b=\frac{1}{9}(\because a>0,b>0)$
And $a+b=\frac{7}{36}$