- Home
- Standard 11
- Mathematics
8. Sequences and Series
hard
The greatest integer less than or equal to the sum of first $100$ terms of the sequence $\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots \ldots$ is equal to
A
$99$
B
$98$
C
$89$
D
$88$
(JEE MAIN-2022)
Solution
$\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots$
$\left(1-\frac{2}{3}\right)+\left(1-\frac{4}{9}\right)+\left(1-\frac{8}{27}\right)+\left(1-\frac{16}{81}\right) \ldots .100 \text { terms }$
$100-\left[\frac{2}{3}+\left(\frac{2}{3}\right)^{2}+\ldots\right]$
$100-\frac{2}{3}\left(1-\left(\frac{2}{3}\right)^{100}\right)$
$100-2\left(1-\left(\frac{2}{3}\right)^{100}\right)$
$S =98+2\left(\frac{2}{3}\right)^{100}$
$\Rightarrow[ S ]=98$
Standard 11
Mathematics
