Question

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

Answer 1

$b-c,2b-x,b-a$ are in HP

$\Rightarrow 2b-x=\frac{2}{(\frac{1}{b}-c+\frac{1}{b}-a)}$ [If $a,b,c$ are in HP then $b=2/(1/a+1/c)]$

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

Claim $a-\frac{x}{2},b-\frac{x}{2},c-\frac{x}{2}$ are in G.P

$(b-\frac{x}{2}{)}^2=(b-b+\frac{(b-c)(b-a)}{(b-a)+(b-c)}{)}^2=\frac{(b-c{)}^2(b-a{)}^2}{\left(\right(b-a)+(b-c){)}^2}$

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

$=[-(b-a)+\frac{(b-c)(b-a)}{(b-a)+(b-c)}][-(b-c)+\frac{(b-c)(b-a)}{(b-a)+(b-c)}]$

$=(b-a)(b-c)[-1+\frac{b-c}{(b-a)+(b-c)}][-1+\frac{(b-a)}{(b-a)+(b-c)}]$

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

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

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

We know that $a,b,c$ are in G.P if $b^2=ac$

$\because {(b-\frac{x}{2})}^2=(a-\frac{x}{2})(c-\frac{x}{2})$

$\Rightarrow (a-\frac{x}{2}),(b-\frac{x}{2}),(c-\frac{x}{2})$ are in G.P