Question

If $b-c,2b-x$ and $b-a$ are in $H.P.$ then $a-\frac{x}{2},b-\frac{x}{2}$ and $c-\frac{x}{2}$are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. None

Answer 1

The correct option is B $G.P.$

$b-c,2b-x$ and $b-a$ are in $H.P.$
$\Rightarrow 2b-x=\frac{2(b-c)(b-a)}{2b-(a+c)}\Rightarrow x=2b-\frac{2(b^2-ab-bc+ac)}{2b-(a+c)}=\frac{2(b^2-ac)}{2b-(a+c)}a-\frac{x}{2}=a-\frac{(b^2-ac)}{2b-(a+c)}=\frac{(a-b{)}^2}{a+c-2b}b-\frac{x}{2}=b-\frac{(b^2-ac)}{2b-(a+c)}=\frac{(a-b)(c-b)}{2b-a-c}c-\frac{x}{2}=c-\frac{(b^2-ac)}{2b-(a+c)}=\frac{(b-c{)}^2}{a+c-2b}\Rightarrow {(b-\frac{x}{2})}^2=(a-\frac{x}{2})(c-\frac{x}{2})$
so $(a-\frac{x}{2}),(b-\frac{x}{2}),(c-\frac{x}{2})$ are in $G.P.$