Question

If a, b and c are in HP, then $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in.

• A. AP
• B. GP
• C. HP
• D. Cannot be determined uniquely

Answer 1

The correct option is C HP

The best way is to assume some HP and then solve it.
Let a, b, c be 10, 12, 15 respectively, which are in HP.
Then, $\frac{a}{b+c}=\frac{10}{12+15}=\frac{10}{27}=0.370$
$\frac{b}{c+a}=\frac{12}{10+15}=\frac{12}{25}=0.480$
$\frac{c}{a+b}=\frac{15}{10+12}=\frac{15}{22}=0.681$
Check for AP, GP and HP.
We find that the numbers are in HP.

Conventional Approach:
a, b, c are in H.P. so, $\frac{1}{a},\frac{1}{b}and\frac{1}{c}$ will be in A.P.
or, $\frac{a+b+c}{a},\frac{a+b+c}{b}and\frac{a+b+c}{c}$ will be in A.P.
or, $1+\frac{b+c}{a},1+\frac{a+c}{b}and1+\frac{a+b}{c}$ will be in A.P.
Hence $\frac{b+c}{a},\frac{a+c}{b}and\frac{a+b}{c}$ are in A.P.
So, $\frac{a}{b+c},\frac{b}{c+a}and\frac{c}{a+b}$ will be in H.P.
Let us take $1,\frac{1}{2},\frac{1}{3}$ (which are in H.P.)
$\frac{a}{b+c}=\frac{6}{5},\frac{b}{a+c}=\frac{3}{8},\frac{c}{a+b}=\frac{2}{9}$
Now, when we check these values of A.P, G.P. and H.P. we find that $\frac{8}{3}$ is the AM of $\frac{5}{6}and\frac{9}{2}$