Question

If sum of first $n$ terms of a sequence (having negative terms) is given by $S_n=\left(1+2 T_n\right)\left(1-T_n\right)$ where $T_n$ is the $n^{\text {th }}$ term of series then $T_2^2=\frac{\sqrt{a}+\sqrt{b}}{4}(a, b \in N)$. Find the value of $(a+b)$.

Answer 1

We have $S_n=\left(1+2 T_n\right)\left(1-T_n\right)$....(1)
Putting $n =1$, in equation (1), we get
$S _1=\left(1+2 T _1\right)\left(1- T _1\right)= T _1 $
$\Rightarrow 1+ T _1-2 T _1^2= T _1 \Rightarrow T _1=\frac{1}{\sqrt{2}} \left(\text { As } T _1>0\right)$
Now putting $n =2$ in equation (1), we get
$S _2=\left(1+2 T _2\right)\left(1- T _2\right) $
$\Rightarrow T _1+ T _2=1+ T _2-2 T _2^2 \Rightarrow 2 T _2^2=1-\frac{1}{\sqrt{2}}$
$\therefore T _2^2=\frac{\sqrt{2}-1}{2 \sqrt{2}}=\frac{\sqrt{4}-\sqrt{2}}{4} \equiv \frac{\sqrt{ a }-\sqrt{ b }}{4} \text { (Given) }$
So, $a=4$ and $b=2$
Hence $(a+b)=4+2=6$