Question

the first term of sequence is 3/8. If 'a' is a next term is 1-a/1+a then the 2007th term is
​

Answer 1

Let $a$ be the $k^{th}$ term of the sequence.

$\longrightarrow t_k=a$

Given that the next term is $\dfrac{1-a}{1+a},$ i.e., $(k+1)^{th}$ term.

$\longrightarrow t_{k+1}=\dfrac{1-a}{1+a}$

Or,

$\longrightarrow t_{k+1}=\dfrac{1-t_k}{1+t_k}$

So, $(k+2)^{th}$ term will be,

$\longrightarrow t_{k+2}=\dfrac{1-t_{k+1}}{1+t_{k+1}}$

$\longrightarrow t_{k+2}=\dfrac{\left(1-\dfrac{1-a}{1+a}\right)}{\left(1+\dfrac{1-a}{1+a}\right)}$

$\longrightarrow t_{k+2}=\dfrac{\left(\dfrac{(1+a)-(1-a)}{1+a}\right)}{\left(\dfrac{(1+a)+(1-a)}{1+a}\right)}$

$\longrightarrow t_{k+2}=\dfrac{1+a-1+a}{1+a+1-a}$

$\longrightarrow t_{k+2}=\dfrac{2a}{2}$

$\longrightarrow t_{k+2}=a$

$\longrightarrow t_{k+2}=t_k$

This means the terms are repeating at alternate positions as,

$\longrightarrow t_1=t_3=t_5=\dots$

$\longrightarrow t_2=t_4=t_6=\dots$

I.e., terms at odd positions are same. And terms at even position are same.

Therefore, we get,

$\longrightarrow\underline{\underline{t_1=t_{2007}=\dfrac{3}{8}}}$

Hence (C) is the answer.