2105/037) If one AM A and two GMs p and q are inserted between two given numbers, show that $\dfrac{p^2}{q} + \dfrac{q^2}{p}= 2A$

because A is AM so 

$2A = (a+b)\cdots(1)$

and p q are 2 GMS

$a,p, q, b$ are in GP

ratio = t so 
$p = at\cdots(2)$
$q = at^2\cdots(3)$
$b = at^3\cdots(4)$

now 
$(\dfrac{p^2}{q} + \dfrac{q^2}{p})$
= $\dfrac{a^2t^2}{at^2} + \dfrac{a^2t^4}{at}$ (from 2 and 3)
= $a + at^3$
= $a + b$ (from (4))
= 2A (from 1)