If one GM, g and two AM's, p and q are inserted between two numbers a and b, then (2 p - q)( p - 2q) is equal to

 Since,  g=$\sqrt{\mathrm{ab}}$also,  a ,  p, q and b are in AP. So. common  difference. d is $\frac{\mathrm{b}-\mathrm{a}}{3}$

$\begin{array}{l}\therefore \mathrm{p}=\mathrm{a}+\mathrm{d}=\mathrm{a}+\frac{\mathrm{b}-\mathrm{a}}{3}=\frac{2\mathrm{a}+\mathrm{b}}{3}\\ \mathrm{q}=\mathrm{b}-\mathrm{d}=\mathrm{b}-\frac{\mathrm{b}-\mathrm{a}}{3}=\frac{\mathrm{a}+2\mathrm{b}}{3}\end{array}$

Now, $\begin{array}{l}(2p-q)(p-2q)\\ =\frac{(4a+2b-a-2b)}{3}\cdot \frac{(2a+b-2a-4b)}{3}\\ =-ab=-g^2\end{array}$