Question

If $A_1,A_2$ be two arithmetic means and $G_1,G_2$ be two geometric means between two positive numbers $a$ and $b,$ then$\left(\frac{A_1+A_2}{G_1{G}_2}\right)$ is equal to

• A. $\frac{ab}{a+b}$
• B. $\frac{2ab}{a+b}$
• C. $\frac{a+b}{2ab}$
• D. $\frac{a+b}{ab}$

Answer 1

The correct option is D $\frac{a+b}{ab}$

Using the formula of A.P.,
$t_n=a+(n–1)d$
$d=\frac{b–a}{3}$ (here $n=4)$
Therefore,$A_1=a+d=a+\frac{b-a}{3}=\frac{2a+b}{3}$
and $A_2=a+2d=a+\frac{2(b–a)}{3}=\frac{a+2b}{3}$
$\therefore A_1+A_2=a+b$

Using the formula of G.P.,
$t_n=a\cdot r^{(n–1)}$
$r=\sqrt[3]{3\frac{b}{a}}$ (here $n=4)$
Therefore,$G_1=a\cdot r=a\sqrt[3]{3{\left(\frac{b}{a}\right)}^2}=(a{b}^2{)}^{\frac{1}{3}}$
and $G_2=a\cdot r^2=a\sqrt[3]{3\frac{b}{a}}=(a^{2}b{)}^{\frac{1}{3}}$
$\therefore G_1\cdot G_2=a\cdot b$

$\therefore \left(\frac{A_1+A_2}{G_1{G}_2}\right)=\frac{a+b}{ab}$