Question

if a,b,c be in hp then prove that a-b/2,b/2,c-b/2 will be in gp

Answer 1

Solution:

  If,a ,b and c are in H.P, then

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

We have to prove that,

                           $a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2}$

                                                 are in G.P.

Suppose , these three numbers are in G.P.

       $[\frac{b}{2}]^2=(a-\frac{b}{2})(c-\frac{b}{2})\\\\ \frac{b^2}{4}=ac-\frac{ab}{2}-\frac{bc}{2}+ \frac{b^2}{4}\\\\ac-\frac{ab}{2}-\frac{bc}{2}=0\\\\ac=\frac{ab}{2}+\frac{bc}{2}\\\\2 ac=a b+ bc\\\\\frac{2}{b}=\frac{a+c}{ac}\\\\\frac{2}{b}=\frac{a}{ac}+\frac{c}{ac}\\\\\frac{2}{b}=\frac{1}{c}+\frac{1}{a}$

which are in H.P.

 Hence Proved.