Question

If $a,b,c$ are in GP, $a-b,c-a,b-c$ are in HP, then the value of $a+4b+c$ is

• A. 12
• B. 0
• C. 11
• D. 2

Answer 1

Answer: (B)

0

$a,b,c$ are in GP $b^2=ac$

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

So, their reciprocals are in AP.

So, $\frac{1}{a-b},\frac{1}{c-a},\frac{1}{b-c}$ are in AP.

$(c-a)=\frac{2(a-b)(b-c)}{(a-c)}-{(c-a)}^2=2(ab-b^2-ac+bc)-{(c-a)}^2=2\left(\right(a+c)b-2ac)-c^2-a^2+2ac=2(a+c)b-4ac-c^2-a^2+6ac=2(a+c)b-c^2-a^2-2ac+8ac-2(a+c)b=0-{(c+a)}^2+8ac-2(a+c)b=0{(c+a)}^2+2(a+c)b-8ac=0(c+a)=\frac{-2b\pm \sqrt{4b^2+32ac}}{2}(a+c)=\frac{-2b\pm 6b}{2}=-4ba+4b+c=0$