Question

If the roots of $a{x}^3+b{x}^2+cx+d=0$ are in H.P., then the roots of $d{x}^3-c{x}^2+bx-a=0$ 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 $\alpha ,\beta ,\gamma$ are the roots of $a{x}^3+b{x}^2+cx+d=0$
Replacing $x\to \frac{-1}{x}$
We get
$a{\left(\frac{-1}{x}\right)}^3+b{\left(\frac{-1}{x}\right)}^2+c\left(\frac{-1}{x}\right)+d=0\Rightarrow d{x}^3-c{x}^2+bx-a=0$
Then roots of $d{x}^3-c{x}^2+bx-a=0$ are $\frac{-1}{\alpha },\frac{-1}{\beta },\frac{-1}{\gamma }$
Now as $\alpha ,\beta ,\gamma$ are in H.P
Gives $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma }$ are in A.P
And $\frac{-1}{\alpha },\frac{-1}{\beta },\frac{-1}{\gamma }$ are in A.P