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(A) Given that $a_{1}, a_{2}, a_{3}, \ldots$ is a $G$.$P$. with common ratio $r$.We have $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.Since $a_{3} = a

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$-171$

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(A) Given that $a_{1}, a_{2}, a_{3}, \ldots$ is a $G$.$P$. with common ratio $r$.We have $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.Since $a_{3} = a_{1}r^{2}$ and $a_{4} = a_{2}r^{2}$,we can write $a_{3} + a_{4} = r^{2}(a_{1} + a_{2}) = 16$.Substituting $a_{1} + a_{2} = 4$,we get $r^{2}(4) = 16$,which implies $r^{2} = 4$,so $r = 2$ or $r = -2$.Given $a_{1} + a_{2} = 4$,we have $a_{1}(1 + r) = 4$.If $r = 2$,$a_{1}(3) = 4 \Rightarrow a_{1} = 4/3$ (not possible as $a_{1} If $r = -2$,$a_{1}(1 - 2) = 4 \Rightarrow -a_{1} = 4 \Rightarrow a_{1} = -4$.Now,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 $4 \lambda = -684$,so $\lambda = -171$.

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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