Question

If a,b,c are positive real numbers in A.P and $a^2,b^2,c^2$ are in H.P, then$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$

Answer 1

Given : $a,b,c$ are in A.P

$a^2,b^2,c^2$ are in H.P

then $2b=a+c$ .....(1)

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

i.e. $\frac{1}{b^2}-\frac{1}{a^2}=\frac{1}{c^2}-\frac{1}{b^2}$

$\Rightarrow \frac{a^2-b^2}{a^2{b}^2}=\frac{b^2-c^2}{c^2{b}^2}$

$\Rightarrow \frac{(a+b)(a-b)}{a^2{b}^2}=\frac{(b-c)(b+c)}{c^2{b}^2}$ ...(1)

Now $\because a,b,c$ are in A.P

$\therefore b-a=c-a$

$\therefore$ (1) becomes $\frac{a+b}{a^2}=\frac{b+c}{c^2}$

$\Rightarrow a{c}^2+b{c}^2=a^{2}b+a^{2}c$

$b{c}^2-a^{2}b=0$

$\Rightarrow a{c}^2-a^{2}c=a^{2}b-b{c}^2$

$\Rightarrow ac(c-a)+b(c-a)(c+a)=0$

$\Rightarrow (c-a)[ac+bc+ca]=0$

$\Rightarrow c=a$ or $ab+bc+ca=0$

$\Rightarrow a=b=c$ then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=3$