If the arithmetic mean and geometric
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 $=\frac{t_{p}+t_{0}}{2}=\frac{5}{4} \& \mathrm{G} \cdot \mathrm{M}=\sqrt{t_{p} t_{q}}=1$

$\Rightarrow 16^{2}\left(-\frac{1}{2}\right)^{p+q-2}=1$

$\Rightarrow(-2)^{8}=(-2)^{(p+q-2)}$

$\Rightarrow p+q=10$

 

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