Question

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

• A. $a=b=c$
• B. $2b=3a+c$
• C. $b^2=\sqrt{\frac{ac}{8}}$
• D. none of these

Answer 1

Answer: (A)

Since $a,b,c$ are in $A.P$.
$\therefore b-a=c-b...\left(1\right)$
Since $a^2,b^2,c^2$ are in $H.P$.
$\therefore \frac{1}{b^2}-\frac{1}{a^2}=\frac{1}{c^2}-\frac{1}{b^2}$
$\Rightarrow \frac{a^2-b^2}{a^2{b}^2}=\frac{b^2-c^2}{b^2{c}^2}$
$\Rightarrow \frac{\left(a-b\right)\left(a+b\right)}{a^2}=\frac{\left(b-c\right)\left(b+c\right)}{c^2}$
$\Rightarrow \frac{a+b}{a^2}=\frac{b+c}{c^2}$ [Using $(1)$]
$\Rightarrow a{c}^2+b{c}^2-a^{2}b-a^{2}c=0$
$\Rightarrow a{c}^2-a^{2}c+b{c}^2-b{a}^2=0$
$\Rightarrow ac\left(c-a\right)+b\left(c^2-a^2\right)=0$
$\Rightarrow \left(c-a\right)\left(ac+b\left(c+a\right)\right)=0$
$\Rightarrow \left(c-a\right)\left(ac+bc+ab\right)=0$
$\Rightarrow$ either $c=a$ or $ab+bc+ca=0$
If $c=a$, from $\left(1\right)b-a=a-b$
$\Rightarrow 2a=2b$
$\Rightarrow a=b$. Hence $a=b=c$
$[ab+bc+ca=0\Rightarrow b\left(a+c\right)=-ca$
$\Rightarrow b.2b=-ca\Rightarrow 2b^2=-ca$
not possible if $a,b,c$ are$+ve]$