Question

If $a,b,c$ are in $G.P.$ and $b-c,c-a,a-b$ are in $H.P.$ then $a+b+c=-3(ac{)}^{1/2}$ If A be $A.M.$ and $G$ be the $G.M.$between two numbers, show that numbers are give .

then statement is ?

• A. True
• B. False

Answer 1

The correct option is B False

Given,

a,b,c are in GP

$\therefore b^2=ac$

Also b-c,c-aand a-b are in HP

$\frac{1}{c-a}-\frac{1}{b-c}=\frac{1}{a-b}-\frac{1}{c-a}$

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

On solving this we get,

2b=a+c

means a,b,c are In AP

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

$⟹{\frac{(a+c}{2^2}}^2=ac$

$⟹a^2+b^2+2ac-4ac=0⟹(a-c{)}^2=0\to a=c$

Taking LHS

$(a+b+c)=(a+\frac{a+c}{2}+c)=a+\frac{2a}{2}+a=3a$

Taking RHS $-3(ac{)}^{\frac{1}{2}}$

=-3a

LHS$\ne$RHS