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) $\frac{p^3+q^3}{pq}$

(b) $\frac{p^3-q^3}{pq}$

(c) $\frac{p^2+q^2}{2}$

(d) $\frac{pq}{2}$

Answer 1

(a) $\frac{p^3+q^3}{pq}$

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{positive}\mathrm{numbers}\mathrm{be}a\mathrm{and}b. \\ a,A\mathrm{and}b\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore 2A=a+b\left(\mathrm{i}\right) \\ \mathrm{Also},a,p,q\mathrm{and}b\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore r={\left(\frac{b}{a}\right)}^{\frac{1}{3}} \\ \text{Again},p=ar\mathrm{and}q=a{r}^2.\left(\mathrm{ii}\right) \\ \mathrm{Now},2A=a+b\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ =a+a\left(\frac{b}{a}\right) \\ =a+a{\left({\left(\frac{b}{a}\right)}^{\frac{1}{3}}\right)}^3 \\ =a+a{r}^3 \\ =\frac{{\left(ar\right)}^2}{a{r}^2}+\frac{{\left(a{r}^2\right)}^2}{ar} \\ =\frac{p^2}{q}+\frac{q^2}{p}\left[\mathrm{Using}\left(\mathrm{ii}\right)\right] \\ =\frac{p^3+q^3}{pq} \end{gathered}$$