Question

If the roots of $p{x}^3+q{x}^2+rx+s=0$are in A.P., then the roots of $8p{x}^3+4q{x}^2+2rx+s=0$ are in

• A. A.P.
• B. G.P
• C. H.P.
• D. A.G.P.

Answer 1

The correct option is A A.P.

Let $\alpha ,\beta ,\gamma$ be roots of $p{x}^3+q{x}^2+rx+s=0$
Now replacing $x\to 2x$, we get
$p{\left(2x\right)}^3+q{\left(2x\right)}^2+r\left(2x\right)+s=0\Rightarrow 8p{x}^3+4q{x}^2+2rx+s=0$
So $\frac{\alpha }{2},\frac{\beta }{2},\frac{\gamma }{2}$ are roots of $8p{x}^3+4q{x}^2+2rx+s=0$
Now as $\alpha ,\beta ,\gamma$ are in A.P
then $\frac{\alpha }{2},\frac{\beta }{2},\frac{\gamma }{2}$ are also in A.P