If the arithmetic mean and geometric mean of the $p^{t h}$ and $q^{t h}$ terms of the sequence $- 16, 8, - 4, 2, . . .$ satisfy the equation $4 x^{2} - 9 x + 5 = 0$ , then $p + q$ is equal to

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Solution

Given, $4 x^{2} - 9 x + 5 = 0$
$\Rightarrow (x - 1) (4 x - 5) = 0$
$\Rightarrow A . M = \frac{5}{4}, G . M = 1 (A . M \geq G . M)$
Again, for the series
$- 16, 8, - 4, 2...$
$p^{t h}$ term $t_{p} = - 16 {(\frac{- 1}{2})}^{p - 1}$
$q^{t h}$ term $t_{p} = - 16 {(\frac{- 1}{2})}^{q - 1}$
Now, A.M $= \frac{t_{p} + t_{q}}{2} = \frac{5}{4}$
& G.M $= \sqrt{t_{p} t_{q}} = 1$
$\Rightarrow 16^{2} {(- \frac{1}{2})}^{p + q - 2} = 1$
$\Rightarrow (- 2)^{8} = (- 2)^{p + q - 2}$
$\Rightarrow p + q = 10$

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If the arithmetic mean and geometric mean of the pth and qth terms of the sequence −16,8,−4,2,... satisfy the equation 4x2−9x+5=0, then p+q is equal to

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If the arithmetic mean and geometric mean of the $p^{t h}$ and $q^{t h}$ terms of the sequence $- 16, 8, - 4, 2, . . .$ satisfy the equation $4 x^{2} - 9 x + 5 = 0$ , then $p + q$ is equal to