Question

The value of $\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+....\infty$ equals to

• A. $\frac{1}{(24{)}^2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{24}$
• D. $\frac{1}{(24{)}^3}$

Answer 1

The correct option is C $\frac{1}{24}$

$S=\frac{1}{6\cdot 10},\frac{1}{10\cdot 14}+\frac{1}{1419}+.....\infty$

$S=\frac{1}{4}(\frac{1}{6}-\frac{1}{10})+\frac{1}{4}(\frac{1}{10}-\frac{1}{14})+\frac{1}{4}(\frac{1}{14}-\frac{1}{18})+.....\infty$

Now take $\frac{1}{4}$ common

$S=\frac{1}{4}[\frac{1}{6}-\frac{1}{16}+\frac{1}{10}-\frac{1}{1/4}+\frac{1}{14}-\frac{1}{18}+1,-\infty ]$

$S=\frac{1}{4}\times \frac{1}{6}=\frac{1}{24}$

option $c$ is correct