If `a x^3+b x^2+c x+d` is divisible by `a x^2+c ,t h e na ,b ,c ,d` are in a. A.P. b. G.P. c. H.P. d. none of these

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

B

Explanation

Since `ax^(3)+bx^(2)+cx+d` is divisible by `ax^(2)+c`.
Therefore, when `ax^(3)+bx^(2)+cx+d` is divided by `ax^(2)+c` the remainder should be zero. If `ax^(3)+bx^(2)+cx+d` is divided by
`ax^(2)+c`, then the remainder is `(bc)/(a)-d`.
`:." "(bc)/(a)-d=0rArrbc=adrArr(b)/(a)=(d)/(c)rArr a,b,c,d` are in G.P.