Question

If $a,b,c$ are in A.P then $a(\frac{1}{b}+\frac{1}{c}),b(\frac{1}{c}+\frac{1}{a}),c(\frac{1}{a}+\frac{1}{b})$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $A.G.P$

Answer 1

The correct option is A $A.P.$

Given $a,b,c$ are in AP

Multiplying $a,b,c$ with $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$, we get

$a(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}),b(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}),c(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ are in AP

$a(\frac{1}{b}+\frac{1}{c})+\frac{a}{a},b(\frac{1}{a}+\frac{1}{c})+\frac{b}{b},c(\frac{1}{a}+\frac{1}{b})+\frac{c}{c}$ are in AP

$a(\frac{1}{b}+\frac{1}{c})+1,b(\frac{1}{a}+\frac{1}{c})+1,c(\frac{1}{a}+\frac{1}{b})+1$ are in AP

Subtracting $1$ from each term we get

$a(\frac{1}{b}+\frac{1}{c}),b(\frac{1}{a}+\frac{1}{c}),c(\frac{1}{a}+\frac{1}{b})$ are in AP