Question 68:
If $a_{1}=\frac{1}{2\times5},a_{2}=\frac{1}{5\times8},a_{3}=\frac{1}{8\times11},...,$ then $a_{1}+a_{2}+a_{3}+...+a_{100}$ is
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If $a_{1}=\frac{1}{2\times5},a_{2}=\frac{1}{5\times8},a_{3}=\frac{1}{8\times11},...,$ then $a_{1}+a_{2}+a_{3}+...+a_{100}$ is
- $\frac{25}{151}$
- $\frac{1}{2}$
- $\frac{1}{4}$
- $\frac{111}{55}$
Option: 1
Explanation:
Explanation:
$a_{100} = \frac{1}{ (3\times100 -1) \times (3\times100 + 2)}= \frac{1}{ 299 \times 302}$
$\frac{1}{2\times5} = \frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$
$\frac{1}{5\times8} = \frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$
$\frac{1}{8\times11} = \frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$
....
$\frac{1}{299\times302} = \frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$
Hence $a_{1}+a_{2}+a_{3}+...+a_{100}$ = $\frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$ + $\frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$ + $\frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$ + ... + $\frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$
= $\frac{1}{3} \times (\frac{1}{2} - \frac{1}{302})$
= $\frac{25}{151}$
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