Question

The $n^{th}$ term of the series $1+3+7+13+21+$ is $9901$. The value of $n$ is ................

• A. 900
• B. 99
• C. 100
• D. 90

Answer 1

Answer: (C)

Given, series
$1+3+7+13+21+...$
Also, $t_n=9901$...(i)
Let $S_n=1+3+7+13+21+...n$ terms
and $S_n=1+3+7+13+...n$ terms
On subtracting
$0=(1+2+4+6+8+...)-t_n$
$t_n=1+2+4+6+8+...n$ terms
$t_n=1+2[1+2+3+4+...(n-1)$ terms]
$t_n=1+2\left[\frac{(n-1)(n-1+1)}{2}\right]$
$t_n=1+n(n-1)$
$9901=1+n(n-1)$ [from Eq. (i)]
$n^2-n-9900=0$
$n^2-100n+99n-9900=0$
$n(n-100)+99(n-100)=0$
$(n-100)(n+99)=0$
$\Rightarrow n=100(n=-99$, neglecting)
(because terms not negative)