Question

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

Answer 1

H.P condition,

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

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

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

$\therefore a+b+c=3b$ or $a+b+c=-3b$

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

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