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 . I

English

Gujarati

Hindi

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

A

$-171$

B

$171$

C

$\frac{511}{3}$

D

$-513$

(JEE MAIN-2020)

Solution

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

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

$\Rightarrow \mathrm{r}^{2}=4 \Rightarrow \mathrm{r}=-2 \quad$ 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

Share

Similar Questions

If ${p^{th}},\;{q^{th}},\;{r^{th}}$ and ${s^{th}}$ terms of an $A.P.$ be in $G.P.$, then $(p – q),\;(q – r),\;(r – s)$ will be in

medium

The ${20^{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + …….$ will be

easy

Let ${A_n} = \left( {\frac{3}{4}} \right) – {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} – ….. + {\left( { – 1} \right)^{n – 1}}{\left( {\frac{3}{4}} \right)^n}$ and $B_n \,= 1 – A_n$ . Then, the least odd natural number $p$ , so that ${B_n} > {A_n}$, for all $n \geq p$ is

hard

(JEE MAIN-2018)

Show that the products of the corresponding terms of the sequences $a,$ $ar,$ $a r^{2},$ $……a r^{n-1}$ and $A, A R, A R^{2}, \ldots, A R^{n-1}$ form a $G .P.,$ and find the common ratio.

easy

Find the sum of $n$ terms in the geometric progression $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$

easy