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

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8. Sequences and Series

hard

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 $\sum\limits_{i=1}^{9} a_{i}=4 \lambda,$ then $\lambda$ is equal to

$-171$

$171$

$\frac{511}{3}$

$-513$

(JEE MAIN-2020)

$a_{1}+a_{2}=4$

$\mathrm{r}^{2} \mathrm{a}_{1}+\mathrm{r}^{2} \mathrm{a}_{2}=16$

$\Rightarrow \mathrm{r}^{2}=4 \Rightarrow \mathrm{r}=-2 \quad$ as $\mathrm{a}_{1}<0$

and $a_{1}+a_{2}=4$

$a_{1}+a_{1}(-2)=4 \Rightarrow a_{1}=-4$

$4 \lambda=(-4)\left(\frac{(-2)^{9}-1}{-2-1}\right)=(-4) \times \frac{513}{3}$

$\Rightarrow \lambda=-171$

Standard 11

Mathematics

easy

medium

(JEE MAIN-2020)

easy

easy

easy