Question

The value of $\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdots \frac{99}{100}$
lies in the interval,

• A. $(\frac{1}{15},\frac{1}{10})$
• B. $(\frac{1}{20},\frac{1}{15})$
• C. $(\frac{1}{10},\frac{1}{5})$
• D. $(\frac{1}{10},1)$

Answer 1

The correct option is A $(\frac{1}{15},\frac{1}{10})$

Let $X=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdots \frac{99}{100}$
$\Rightarrow X<\frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7}\cdot \frac{8}{9}\cdots \frac{100}{101}$ as
$\frac{n}{n+1}<\frac{n+1}{n+2}$ for $n>0$
$\Rightarrow X^2<\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdot \frac{4}{5}\cdots \frac{99}{100}\cdot \frac{100}{101}=\frac{1}{101}<\frac{1}{100}$
$\Rightarrow X<\frac{1}{10}$
Also $x><\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdot \frac{4}{5}\cdots \frac{98}{99}\cdot \frac{99}{100}=\frac{1}{200}<\frac{1}{225}$
$\Rightarrow X<\frac{1}{15}$
$\therefore \frac{1}{15}<X<\frac{1}{10}\Rightarrow X\in (\frac{1}{15},\frac{1}{10})$