If p , g_1 , g_2 and q are in GP and m is the arithmetic mean of p...

Given $p$, $g_1$, $g_2$, $q$ are in GP.

Let the common ratio of the GP be $r$.

Then $g_1 = pr$, $g_2 = pr^2$, and $q = pr^3$.

The arithmetic mean of $p$ and $q$ is $m$, so $m = \dfrac{p + q}{2} \Rightarrow p + q = 2m$.

Consider the expression $\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1}$.

Substituting the values of $g_1$ and $g_2$:

$\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1} = \dfrac{(pr)^2}{pr^2} + \dfrac{(pr^2)^2}{pr}$

$= \dfrac{p^2r^2}{pr^2} + \dfrac{p^2r^4}{pr}$

$= p + pr^3$

Since $q = pr^3$, we get:

$= p + q$

Substituting $p + q = 2m$:

$\dfrac{g_1^2}{g_2} + \dfrac{g_2^2}{g_1} = 2m$

Answer: $2m$