Question

If the roots of $a{x}^3+b{x}^2+cx+d=0$ are in $G.P.$, then the roots of $a{x}^3+3b{x}^2+9cx+27d=0$ are in

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

Answer 1

Answer: (B)

G.P.

$a{x}^3+3b{x}^2+9cx+27d=0$

$⟹a(\frac{x}{3}{)}^3+b(\frac{x}{3}{)}^2+c\left(\frac{x}{3}\right)+d=0$

So, if $\alpha ,\beta ,\gamma$ are the roots of $a{x}^3+b{x}^2+cx+d=0$

Then, the roots of $a{x}^3+3b{x}^2+9cx+17d=0$ are $3\alpha ,3\beta ,3\gamma$

So, if $\alpha ,\beta ,\gamma$ are in G.P then $3\alpha ,3\beta ,3\gamma$ are also in G.P

So, the roots of $a{x}^3+3b{x}^2+9cx+17d=0$ are also in G.P.

So, the answer is option (B)