Question

If $A_1,A_2$ be two $A{M}^{\prime }s$ and $G_1,G_2$ be two $G{M}^{\prime }s$ between two numbers $a$ and $b$, then $\frac{A_1+A_2}{G_1{G}_2}$ is equal to

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

Answer 1

Answer: (C)

$\frac{a+b}{ab}$

As per question

$AP$ series will be $a,A_1,A_2,b$

$d=\frac{b-a}{3}$

$A_1=a+\frac{b-a}{3}=(2a+b)/3$

$A_2=a+\frac{2(b-a)}{3}=(a+2b)/3$

$GP$ series will be $a,G_1,G_2,b$

$r=(\frac{b}{a}{)}^{1/3}$

$G_1=a(\frac{b}{a}{)}^{1/3}=b^{1/3}{a}^2/3$

$G_2=a(\frac{b}{a}{)}^{2/3}=b^{2/3}{a}^1/3$

Required ratio will be $\frac{A_1+A_2}{G_1{G}_2}=\frac{\frac{2a+b}{3}+\frac{a+2b}{3}}{a^{1/3}{b}^{2/3}{a}^{2/3}{b}^{1/3}}=\frac{a+b}{ab}$