Question

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

• A. $a=b=c$
• B. $a^2=b^2=\frac{c^2}{2}$
• C. $a,b,c$ are in $G.P.$
• D. $\frac{-a}{2},b,c$ are in $G.P.$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $a=b=c$
C $a,b,c$ are in $G.P.$
D $\frac{-a}{2},b,c$ are in $G.P.$

We have $2b=a+c$ and $b^2=\frac{2a^2{c}^2}{a^2+c^2}.....\left(i\right)$
On eliminating b, we get
$8a^2{c}^2=(a^2+c^2+2ac)(a^2+c^2)$
Which can be arranged as
$(a^2+c^2-2ac)(a^2+c^2+4ac)=0$
$\Rightarrow eithera=c$ or $(a+c{)}^2+2ac=0$
If $a=c$ then $a=b=c$
$\Rightarrow$ a, b, c may be treated as three numbers in G .P.
If $(a+c{)}^2+2ac=0$, then by using (i) choice (D) follows.