Question:
If A be one A.M. and p, q be two G.M.'s between two numbers, then 2 A is equal to
(a) $\frac{p^{3}+q^{3}}{p q}$
(b) $\frac{p^{3}-q^{3}}{p q}$
(c) $\frac{p^{2}+q^{2}}{2}$
(d) $\frac{p q}{2}$
Solution:
(a) $\frac{p^{3}+q^{3}}{p q}$
Let the two positive numbers be $a$ and $b$.
$a, A$ and $b$ are in A.P.
$\therefore 2 A=a+b$ ...(i)
Also, $a, p, q$ and $b$ are in G.P.
$\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{3}}$
Again, $p=a r$ and $q=a r^{2}$ ...(ii)
Now, $2 A=a+b \quad[$ From (i) $]$
$=a+a\left(\frac{b}{a}\right)$
$=a+a\left(\left(\frac{b}{a}\right)^{\frac{1}{3}}\right)^{3}$
$=a+a r^{3}$
$=\frac{(a r)^{2}}{a r^{2}}+\frac{\left(a r^{2}\right)^{2}}{a r}$
$=\frac{p^{2}}{q}+\frac{q^{2}}{p} \quad[\mathrm{U} \sin g(\mathrm{ii})]$
$=\frac{p^{3}+q^{3}}{p q}$