Question

If `ax^3 + bx^2 + cx + d` is divisible by `ax^2 + c`, then `a, b, c, d` are in (a) AP (b) GP (c) HP
A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer 1

Correct Answer - B
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.

Answer 2

The answer is B it is an G.P