Question

If a, b, c are real and in A.P. and $a^2,b^2,c^2$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

The correct option is A $a=b=c$

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

So, $b-a=c-b$ ......(i)

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

So, $\frac{1}{b^2}-\frac{1}{a^2}=\frac{1}{c^2}-\frac{1}{b^2}$

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

$a{c}^2+b{c}^2=a^{2}b+a^{2}c$ ..[Since,$b-a=c-b$ from (i)]

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

Either $c-a=0$ or $ab+bc+ca=0$

So, $c=a$

From (i), we have

$b=a$

Thus, $a=b=c$