Suppose a, b are two positive numbers. The product of three arithmetic means inserted between$a, b$is $15/2$ and product of three arithmetic means inserted between 1/a and $1/b$ is $5/18.$Then ab is equal to:

Let three arithmetic means between a and b be $a_1, a_2, a_3$. Then $a, a_1, a_2, a_3, b$are in A.P. 

Let common difference of this A.P. be d. Then 

$\begin{array}{l}d=\frac{1}{4}(b-a)\\ a_1=a+b=\frac{1}{4}(3a+b)\\ a_2=a+2d=\frac{1}{2}(a+b)\\ a_3=a+3d=\frac{1}{4}(a+3b)\end{array}$

If $A_1, A_2, A_3$ are arithmetic means between $1/a and 1/b$, then

$\begin{array}{r}{A}_1=\frac{1}{4}\left(\frac{3}{a}+\frac{1}{b}\right)=\frac{a+3b}{4ab}\\ A_2=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{a+b}{2ab}\end{array}$

$A_3=\frac{1}{4}\left(\frac{1}{a}+\frac{3}{b}\right)=\frac{3a+b}{4ab}$

Now, $\frac{1}{64}(3a+b)(a+b)(a+3b)=\frac{15}{2}$

and $\frac{1}{64(ab{)}^3}(a+3b)(a+b)(3a+b)=\frac{5}{18}$

Thus, $(ab{)}^3=\frac{15/2}{5/18}=27$

$\Rightarrow ab=3$