Question

Let $\alpha, \beta, \gamma, \delta$ are zeroes of $P(x)=5 x^{4}+p x^{3}+q x^{2}+r x+$ $s(p, q, r, s \in R)$ and $\alpha, \gamma, \delta$ are zeroes of $Q(x)=x^{3}-9 x^{2}$ $+a x-24(\alpha < \beta < \gamma < \delta) .$ If $\alpha, \gamma, \delta$ (taken in that order) are in arithmetic progression and $\alpha, \beta, \gamma, \delta$ (taken in that order) are in harmonic progression, then find the value of $\left|\frac{P(1)}{Q(1)}\right|$.

Answer 1

$2 \gamma=\alpha+\delta$ and $\alpha+\beta+\gamma=9$
$ \Rightarrow \gamma=3$
$\alpha \gamma \delta=24 \Rightarrow \gamma=3, \delta=4, \alpha=2$
$\because \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$ are in A.P.
$\Rightarrow \beta=\frac{12}{5}$
$\left|\frac{P(x)}{Q(x)}\right|=|5(x-\beta)|=7$