Question

if $a+b+c=6$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}$, then find $\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}$

• A. $6$
• B. $4$
• C. $9$
• D. $12$

Answer 1

The correct option is A $6$

We have,

$a+b+c=6......\left(1\right)$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}.......\left(2\right)$

On multiplying (1) and ( 2) and to and we get,

$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=6\times \frac{3}{2}$

$\Rightarrow (\frac{a}{a}+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{b}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+\frac{c}{c})=9$

$\Rightarrow (1+1+1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c})=9$

$\Rightarrow \frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}=9-3$

$\Rightarrow \frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}=6$

$\Rightarrow \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=6$

Hence, this is the answer.