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(A) Given $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.Since it is a $G$.$P$.,$a_{2} = a_{1}r$ and $a_{3} = a_{1}r^{2}$,$a_{4} = a_{1}r^{3}$.$a_{1}(1

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$-171$

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(A) Given $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.Since it is a $G$.$P$.,$a_{2} = a_{1}r$ and $a_{3} = a_{1}r^{2}$,$a_{4} = a_{1}r^{3}$.$a_{1}(1 + r) = 4$  --- $(1)$$a_{1}r^{2}(1 + r) = 16$ --- $(2)$Dividing $(2)$ by $(1)$,we get $r^{2} = 4$,so $r = 2$ or $r = -2$.If $r = 2$,$a_{1}(1 + 2) = 4 \Rightarrow a_{1} = 4/3 > 0$,which contradicts $a_{1} Thus,$r = -2$. Substituting in $(1)$,$a_{1}(1 - 2) = 4$ $\Rightarrow -a_{1} = 4$ $\Rightarrow a_{1} = -4$.The sum of the first $9$ terms is $S_{9} = a_{1} \frac{r^{9} - 1}{r - 1} = (-4) \frac{(-2)^{9} - 1}{-2 - 1} = (-4) \frac{-512 - 1}{-3} = (-4) \frac{-513}{-3} = (-4) 	imes 171 = -684$.Given $S_{9} = 4 \lambda$,we have $4 \lambda = -684 \Rightarrow \lambda = -171$.

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a $G$.$P$. such that $a_{1} < 0$; $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$. If $\sum_{i=1}^{9} a_{i} = 4 \lambda$,then $\lambda$ is equal to:

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