Let $a_1,a_2,\dots ,a_9$ be in G.P. with $a_1<0$ such that

$a_1+a_2=4,a_3+a_4=16.$ If $\sum _{i=1}^9 a_i=4\lambda$ then $\lambda$ is equal to

Let $r$be the common ratio of the given G.P. It is

given that

$\begin{array}{l}{a}_1+a_2=4\text{ and }{a}_3+a_4=16\\ \Rightarrow a_1+a_{1}r=4\text{ and }{a}_1{r}^2+a_1{r}^3=16\\ \Rightarrow a_1(1+r)=4\text{ and }{a}_1{r}^2(1+r)=16\\ \Rightarrow r=\pm 2\end{array}$

When $r=2$

$a_1(1+r)=4\Rightarrow a_1=\frac{4}{3}$

But, it is given that $a_1<0$.So $r=2$ is not possible

When $r=-2$

$\begin{array}{l}{a}_1(1+r)=4\Rightarrow a_1=-4\\ \therefore \sum _{i=1}^9 a_i=4\lambda \\ \Rightarrow -4\left\{\frac{(-2{)}^9-1}{-2-1}\right\}=4\lambda \Rightarrow \lambda =\frac{-513}{3}=-171\end{array}$