Question

If a, b, c are in G.P. and b-c, c-a, a-b are in H.P. then prove that a+b+c =${-3\left(ac\right)}_2^1$

Answer 1

$b-c$, $c-a$, $a-b$ are in $HP$

$\Rightarrow \frac{1}{a-b}+\frac{1}{b-c}=\frac{2}{c-a}$

Clearing the denominations, we get

$a^2+c^2-2b^2+2ab+2bc-4ac=0$

This gives us

${(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ac=0+3b^2+6ac={9b}^2$

Hence either $a+b+c=3b$ or $a+b+c=-3b$

Now in the first case $a+c=2b$ and $ac=b^2$, Hence we must have $a=b=c$. This means that $a-b$, $c-a$, $b-c$ is not a harmonic progression.

Thus, We must have $a+b+c=-3b$

Now given $abc$ are in $GP$

$\Rightarrow b^2=ac$

$\Rightarrow b={ac}^{1/2}$

$\therefore$ $a+b+c=-3b=-3{\left(ac\right)}^{1/2}$

Hence proved.