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(C) The given sequence is a geometric progression with first term $a = -16$ and common ratio $r = -1/2$.The $n^{	ext{th}}$ term is $t_{n} = a r^{n-1}

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$10$

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(C) The given sequence is a geometric progression with first term $a = -16$ and common ratio $r = -1/2$.The $n^{	ext{th}}$ term is $t_{n} = a r^{n-1} = -16(-1/2)^{n-1}$.Let the arithmetic mean and geometric mean of $t_{p}$ and $t_{q}$ be $x_{1}$ and $x_{2}$. The equation $4x^{2}-9x+5=0$ has roots $x = 1$ and $x = 5/4$.Since the arithmetic mean is greater than the geometric mean for distinct terms,we have $AM = 5/4$ and $GM = 1$.$AM = (t_{p} + t_{q})/2 = 5/4 \Rightarrow t_{p} + t_{q} = 5/2$.$GM = \sqrt{t_{p} t_{q}} = 1 \Rightarrow t_{p} t_{q} = 1$.Substituting $t_{p} = -16(-1/2)^{p-1}$ and $t_{q} = -16(-1/2)^{q-1}$:$t_{p} t_{q} = 256(-1/2)^{p+q-2} = 1 \Rightarrow (-1/2)^{p+q-2} = 1/256 = (1/2)^{8}$.Since $(-1/2)^{p+q-2} = (1/2)^{8}$,$p+q-2$ must be an even number equal to $8$.$p+q-2 = 8 \Rightarrow p+q = 10$.

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

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