Question

If the angles $A,B,C$ of a triangle are in AP and the sides $a,b,c$ opposite to these angles are in GP, then $a^2,b^2,c^2$ are in

• A. AP
• B. GP
• C. Neither AP nor GP
• D. HP

Answer 1

Answer: (A)

AP

Since, $A,B$ and $C$ are in AP.
Therefore, $B={60}^o$ and as $a,b,c$ are in GP, therefore $b^2=ac$
Consider, $\cos B=\frac{a^2+c^2-b^2}{2ac}$
$\Rightarrow \frac{1}{2}=\frac{a^2+c^2-b^2}{2b^2}$
$\Rightarrow b^2=a^2+c^2-b^2$
$\Rightarrow 2b^2=a^2+c^2$