Question

If $a,b,c$ are in $G.P$ $a-b,c-a,b-c$ are in $H.P$ then the value of $a+4b+c$ is equal to

Answer 1

The condition for a harmonic progression is that

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

clearing denominators$,$ we get

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

The given us$:$

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

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

Now$,$ in the first case$,$

Since, $a+c=2b$ and $ac=b^2,$

Hence$,$ we must have $a=b=c.$ this means that $a-n,c-a,b-c$ is not a harmonic progression$.$

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

So$,$ $a+4b+c=0.$

Hence, this is the answer.