Question

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

Answer 1

Answer: 6

We have $S_n=\left(1+2T_n\right)\left(1−T_n\right)$....(1)
Putting $n=1$, in equation (1), we get
$S_1=\left(1+2T_1\right)\left(1−T_1\right)=T_1$
$⇒1+T_1−2T_1^2=T_1⇒T_1=\frac{1}{\sqrt{2}}\left(\text{ As }{T}_1>0\right)$
Now putting $n=2$ in equation (1), we get
$S_2=\left(1+2T_2\right)\left(1−T_2\right)$
$⇒T_1+T_2=1+T_2−2T_2^2⇒2T_2^2=1−\frac{1}{\sqrt{2}}$
$∴T_2^2=\frac{\sqrt{2}−1}{2\sqrt{2}}=\frac{\sqrt{4}−\sqrt{2}}{4}≡\frac{\sqrt{a}−\sqrt{b}}{4}\text{ (Given) }$
So, $a=4$ and $b=2$
Hence $(a+b)=4+2=6$