If the first three terms of a sequence $\frac{1}{16},a,b,\frac{1}{6}$ are in G.P. and the last three are in H.P., then the values of a and b respectively are

$\frac{1}{16},a,b\text{ are in G.P. }$

$\Rightarrow a^2=b/16$              (1)

Also, as $a, b, 1/6$ are in H.P

$b=\frac{2\left(a\right)(1/6)}{a+1/6}=\frac{2a}{6a+1}$     (2)

From (1) and (2)

$16a^2=\frac{2a}{6a+1}$

$\Rightarrow 8a(6a+1)=1[\because a\ne 0]$

$\begin{array}{l}\Rightarrow 48a^2+8a-1=0\\ \Rightarrow (12a-1)(4a+1)=0\\ \Rightarrow a=1/12,a=-1/4\end{array}$

When $a=1/12,b=1/9$

When $a=-1/4,b=1$