Question

Let $a_{1}, a_{2}, a_{3}, ...$ be a G. P. such that $a_1 < 0$, $a_1 +a_2 = 4$ and $a_3 + a_4 = 16.$ If $\displaystyle \sum_{i=1}^9a_i=4\lambda$, then $\lambda$ is equal to :

• A. $171$
• B. $\frac{511}{3}$
• C. $-171$
• D. $513$

Answer 1

Answer: (C)

$a_{1}+a_{2}=4$
$r^{2}a_{1}+r^{2}a_{2}=16$
$\Rightarrow r^{2}=4 \Rightarrow r=-2\,as \,a_{1} < 0$
and $a_{1}+a_{2}=4$
$a_{1}+a_{2}\left(-2\right)=4 \Rightarrow a_{1}=-4$
$4\lambda=\left(-4\right)\left(\frac{\left(-2\right)^{9}-1}{-2-1}\right)=\left(-4\right)\times\frac{513}{3}$
$\Rightarrow \lambda=-171$