<p>If \(ax^3 + bx^2 + cx + d\) is divisible by \(ax^2 + c\), then \(a, b, c, d\) are in</p>
Step-by-Step Solution
Key Concept: Use the divisibility condition to set the remainder to zero, which yields the GP relation $bc = ad$.
<p><strong>Solution:</strong></p><p>Since $ax^3 + bx^2 + cx + d$ is divisible by $ax^2 + c$, the remainder must be zero when dividing.</p><p>By polynomial division, the remainder is $d - \frac{bc}{a}$.</p><p>For divisibility: $d - \frac{bc}{a} = 0$</p><p>$bc = ad$</p><p>$\frac{b}{a} = \frac{d}{c}$</p><p>This is the condition for $a, b, c, d$ to be in GP.</p><p>∴ Answer is (b)</p>
Correct Answer: b