Question

If $a,b,c$ are in A.P., $a^2,b^2,c^2$ are in H.P., then prove that either $a=b=c$ or $a,b,-$ $\frac{c}{2}$ form a G.P.

Answer 1

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

$\Rightarrow 2b=a+c$……. (1)

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

$\frac{1}{b^2}-\frac{1}{d^2}=\frac{1}{c^2}-\frac{1}{b^2}$

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

$a{c}^2+b{c}^2=a^{2}b+a^{2}c[\because a–b=b–c]$

$ac(c–a)+b(c–a)(c+a)=0$

$(c–a)(ab+bc+ca)=0$

either $c–a=0orab+bc+ca=0$

either $c=a$ or $(a+c)b+ca=0$ and then form (i) $2b^2+ca=0$

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

i.e. $a,b,\frac{-c}{2}$ are in G. P. Hence Proved