Question

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

• A. $\text{A}\text{.P}$
• B. $\text{G}\text{.P}$
• C. $\text{H}\text{.P}$
• D. None

Answer 1

The correct option is B $\text{G}\text{.P}$

Given, $b-c,2bx,b-a$ are in $HP$

$\Rightarrow 2b-x=2/(1/b-c+1/b-a)$ [if $a,b,c$ are in $HP$ then $b=2/(1/a+1/c)$

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

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

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

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

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

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

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

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

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

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

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

$\because {(b-\frac{x}{2})}^2=(a-\frac{x}{2})(c-\frac{x}{1})$

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