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:
Correct Option: , 2
Since, $a_{1}+a_{2}=4 \Rightarrow a_{1}+a_{1} r=4$ ...(i)
$a_{3}+a_{4}=16 \Rightarrow a_{1} r^{2}+a_{1} r^{3}=16$...(ii)
From eqn. (i), $a_{1}=\frac{4}{1+r}$ and substituting the value of
$a_{1}$, in eqn (ii),
$\left(\frac{4}{1+r}\right)^{r^{2}}+\left(\frac{4}{1+r}\right)^{r^{3}}=16$
$\Rightarrow 4 r^{2}(1+r)=16(1+r)$
$\Rightarrow r=4 \quad \therefore \quad r=\pm 2$
$r=2, a_{1}(1+2)=4 \Rightarrow a_{1}=\frac{4}{3}$
$r=-2, a_{1}(1-2)=4 \Rightarrow a_{1}=-4$
$\sum_{i=1}^{a} a_{i}=\frac{a_{1}\left(r^{q}-1\right)}{r-1}=\frac{(-4)\left((-2)^{9}-1\right)}{-2-1}$
$=\frac{4}{3}(-513)=4 \lambda \quad \Rightarrow \quad \lambda=-171$
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