Question

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

Answer 1

Finding the value of $a+4b+c$:

Given $a,b,c$ are in $G.P$ so $b^2=ac...\left(i\right)$

and $a-b,c-a,b-c$ are in $H.P$

so $\frac{1}{(a-b)},\frac{1}{(c-a)},\frac{1}{(b-c)}$are in $A.P$

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

$\Rightarrow$ $\frac{2}{(c-a)}=\frac{\left[\right(b-c)+(a-b\left)\right]}{(a-b)(b-c)}$

$\Rightarrow$ $\frac{2}{(c-a)}=\frac{(a-c)}{(a-b)(b-c)}$

$\Rightarrow$ $2ab–2b^2–2ac+2bc=-c^2-a^2+2ac$

$\Rightarrow$$a^2+b^2+c^2+2ab+2bc+2ac=3b^2+6ac$

$\Rightarrow$ ${(a+b+c)}^2=3b^2+6ac$

$\Rightarrow$ ${(a+b+c)}^2=3b^2+6b^2(from(i\left)\right)$

$\Rightarrow$ ${(a+b+c)}^2=9b^2$

$\Rightarrow$ $(a+b+c)=\pm 3b$

When $(a+b+c)=-3b$

$\Rightarrow$ $a+b+c+3b=0$

$\Rightarrow$ $a+4b+c=0$

Hence, the value of $a+4b+c$ is $0$