If the arithmetic mean and the geometric mean of the ^ th and q^ th terms of the sequence -16,8,-4,2,… satisfy the equation 4x^2-9x+5=0, then p+q is…

Let {t}_{p}&{t}_{q} is the {p}^{th}&{q}^{th}terms of the series $-16,8,-4,2,...$−16,8,−4,2,...

Here $a=-16,r=-\frac{1}{2}$a=−16,r=−21​

Then, ${t}_{p}=-16{(-\frac{1}{2})}^{p-1}$tp​=−16(−21​)p−1

$${t}_{q}=-16{(-\frac{1}{2})}^{q-1}$$tq​=−16(−21​)q−1

The roots of the given quadratic equation $4{x}^{2}-9x+5=0$4x2−9x+5=0 is $x=1,\frac{5}{4}.$x=1,45​.

$$AM=\frac{5}{4},GM=1(\because AM\geq GM)$$AM=45​,GM=1(∵AM≥GM)

Now, given $\frac{5}{4}=\frac{{t}_{p}+{t}_{q}}{2},1={t}_{p}{t}_{q}$45​=2tp​+tq​​,1=tp​tq​

So, $1=256{(-\frac{1}{2})}^{p+q-2}\Rightarrow {2}^{p+q-2}={(-1)}^{p+q-2}{2}^{8}$1=256(−21​)p+q−2⇒2p+q−2=(−1)p+q−228

$$\Rightarrow {2}^{p+q-10}={(-1)}^{p+q-2}$$⇒2p+q−10=(−1)p+q−2

It is equal when ${2}^{p+q-10}=1={(-1)}^{p+q-2}$2p+q−10=1=(−1)p+q−2

Hence, $p+q-10=0\Rightarrow p+q=10$p+q−10=0⇒p+q=10