Question

If the roots of $a{x}^3+b{x}^2+cx+d=0$ are in A.P., then the roots of $a(x+k{)}^3+b(x+k{)}^2+c(x+k)+d=0$ are in

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

Answer 1

Answer: (A)

A.P.

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

Then replacing by $x\to x+k$, we get

$a(x+k{)}^3+b(x+k{)}^2+c(x+k)+d=0$

and roots of this are $\alpha -k,\beta -k,\gamma -k$

Now as $\alpha ,\beta ,\gamma$ are in A.P

Then $\alpha -k,\beta -k,\gamma -k$ are also in A.P