Question:
If the arithmetic mean and geometric mean of the $p^{\text {th }}$ and $q^{\text {th }}$ terms of the sequence $-16,8,-4,2, \ldots \ldots \ldots \ldots$ satisfy the equation $4 x^{2}-9 x+5=0$, then $\mathrm{p}+\mathrm{q}$ is equal to
Solution:
Given, $4 x^{2}-9 x+5=0$
$\Rightarrow(x-1)(4 x-5)=0$
$\Rightarrow$ A. $M=\frac{5}{4}$, G. $M=1 \quad$ (QA. $M>$ G. $M$ )
Again, for the series
$-16,8,-4,2 \ldots \ldots$
$p^{\text {th }}$ term $t_{p}=-16\left(\frac{-1}{2}\right)^{p-1}$
$q^{\text {th }}$ term $t_{p}=-16\left(\frac{-1}{2}\right)^{q-1}$
NoW, A. M $=\frac{t_{p}+t_{0}}{2}=\frac{5}{4} \& \mathrm{G} \cdot \mathrm{M}=\sqrt{t_{p} t_{q}}=1$
$\Rightarrow 16^{2}\left(-\frac{1}{2}\right)^{p+q-2}=1$
$\Rightarrow(-2)^{8}=(-2)^{(p+q-2)}$
$\Rightarrow p+q=10$