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# CAT 2017 [slot 2] Question with solution 34

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
1. $\frac{25}{151}$
2. $\frac{1}{2}$
3. $\frac{1}{4}$
4. $\frac{111}{55}$
Option: 1
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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