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)

$a = b = c$

Answer 2

Since $a,b, c$ are in $A.P$. $\therefore b-a = c-b \quad...\left(1\right)$ Since $a^{2}, b^{2}, c^{2}$ are in $H.P$. $\therefore \frac{1}{b^{2}} -\frac{1}{a^{2}} = \frac{1}{c^{2}}-\frac{1}{b^{2}}$ $\Rightarrow\frac{ a^{2}-b^{2}}{a^{2}b^{2}} =\frac{ b^{2}-c^{2}}{b^{2}c^{2}}$ $\Rightarrow \frac{\left(a-b\right)\left(a+b\right)}{a^{2}} = \frac{\left(b-c\right)\left(b+c\right)}{c^{2}}$ $\Rightarrow \frac{a+b}{a^{2}} = \frac{b+c}{c^{2}} \quad$ [Using $(1)$] $\Rightarrow ac^{2} +bc^{2}-a^{2}b-a^{2}c= 0$ $\Rightarrow ac^{2}-a^{2}c+bc^{2}-ba^{2}=0$ $\Rightarrow ac\left(c-a\right)+b\left(c^{2}-a^{2}\right) = 0$ $\Rightarrow \left(c-a\right)\left(ac+b\left(c+a\right)\right) = 0$ $\Rightarrow \left(c-a\right)\left(ac+bc+ab\right) =0$ $\Rightarrow$ 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 $a=b=c$ $[ab+bc+ca=0 \Rightarrow b\left(a+c\right)= -ca$ $\Rightarrow b.2b= -ca \Rightarrow2b^{2} = -ca$ not possible if $a, b, c$ are$+ve]$