Question

Let $a_{1,}{a}_2,a_3,..$ be terms of a GP. 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

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

Answer 1

The correct option is C

$-171$

Explanation for Correct answer:

Step 1: Finding the value of $\lambda$:

Given $a_{1,}{a}_2,a_3,..$ are terms of a GP.

A sequence, in which each of its terms can be obtained by multiplying or dividing its preceding term by a fixed quantity, is called a Geometric Progression.

$n^{th}$ term of an G.P. is $a_n=a_1{r}^{n-1},\text{where}{a}_1=\text{first term and}r=\text{common ratio}$

Given,

$$\begin{gathered} a_1+a_2=4 \\ \Rightarrow a+ar=4 \\ \Rightarrow a(1+r)=4\to \left(i\right) \\ {a}_3+a_4=16 \\ \Rightarrow a{r}^2+a{r}^3=16 \\ \Rightarrow a{r}^2(1+r)=16\to \left(ii\right) \end{gathered}$$

equation $\left(ii\right)÷\left(i\right)$, we get

$\frac{a{r}^2\left(1+r\right)}{a\left(1+r\right)}=\frac{16}{4}$

$$\begin{gathered} \Rightarrow r^2=4 \\ \Rightarrow r=\pm 2 \end{gathered}$$

Substitute $r=2$in equation $\left(i\right)$

$\Rightarrow a_1=\frac{4}{3}$ is not possible $\because a_1<0$

So $r=-2\Rightarrow a_1=-4$ which is possible

Step 2: Finding the value of $\lambda$:

Sum of $n$ terms of GP

$S_n=\frac{a\left(r^n-1\right)}{\left(r-1\right)}$ where $a_1=\text{first term},r=\text{common ratio}$.

$$\begin{gathered} S_9=\frac{-4\left({\left(-2\right)}^9-1\right)}{\left(-2-1\right)}\left[\because r=-2,a_1=-4\right] \\ {S}_9=\frac{4\left(-512-1\right)}{3} \\ =-171\times 4 \\ \sum _{i=1}^9{a}_i=4\lambda \\ \Rightarrow \left(-171\times 4\right)=4\lambda \\ \Rightarrow \lambda =-171 \end{gathered}$$

Hence, option (C) is the correct answer.