Question

Let $S=\left\{\lambda \mid \sqrt{\frac{11\lambda -2{\lambda }^2-8}{2\lambda }}\in I,\lambda >0\right\}$. All the elements of set $S$ are the roots of the equation $x^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\dots \dots +a_n=0$ where $n$ denotes the number of elements in $S$.
If ( $n-2)$ is the $3^{\text{rd }}$ term of a G.P. such that sum of its infinite term is equal to $a_n$ then its first term is equal to

• A. $a_n$
• B. $\sqrt{a_n}$
• C. $2\sqrt{a_n}$
• D. $\sqrt[4]{a_n}$

Answer 1

Answer: (C)

Now $n-2=4-2=2=a{r}^2$...(1)
and $\frac{a}{1-r}=a_4=16$...(2)
(1) and (2) $\Rightarrow r=\frac{1}{2}$ and $a=8$
$\therefore a=8=2\sqrt{a_n}=2\sqrt{a_4}$