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Question 34 : If a_{1} = \\frac{1}{2 × 5}) , a_{2} = \\frac{1}{5 × 8}) , a_{3} = \\frac{1}{8 × 11}),...., then a_{1} + a_{2} + a_{3} + ...... + a_{100} is

- \\frac{25}{151})
- \\frac{1}{2})
- \\frac{1}{4})
- \\frac{111}{55})

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Given that a_{1} = \\frac{1}{2 × 5}) , a_{2} = \\frac{1}{5 × 8}) , a_{3} = \\frac{1}{8 × 11}),....,

We have to find a_{1} + a_{2} + a_{3} + ...... + a_{100}

This can be done by partial fractions

a_{n} = \\frac{1}{(3𝑛 − 1)(3𝑛 + 2)})

a_{1} = 3 - 1 = 2 , a_{2} = 6 - 1 = 5 , a_{3} = 9 – 1 = 8.

100^{th} term = \\frac{1}{((3 × 100) − 1)((3 × 100) + 2)})

a_{1} = \\frac{1}{2}) - \\frac{1}{5}) = \\frac{5 - 2}{2 × 5}) = \\frac{3}{2 × 5})

a_{2} = \\frac{1}{5}) - \\frac{1}{8}) = \\frac{8 - 5}{5 × 8}) = \\frac{3}{5 × 8})

But these are not of the form \\frac{1}{(3𝑛 − 1)(3𝑛 + 2)}). Hence these are written as,

\\frac{1}{3})[\\frac{1}{2}) - \\frac{1}{5})] = \\frac{5 - 2}{2 × 5}) = \\frac{3}{2 × 5}) × \\frac{1}{3}) = \\frac{1}{2 × 5})

\\frac{1}{3})[\\frac{1}{5}) - \\frac{1}{8})] = \\frac{8 - 5}{5 × 8}) = \\frac{3}{5 × 8}) × \\frac{1}{3}) = \\frac{1}{5 × 8})

Hence this can be written as,

\\frac{1}{3})[(\\frac{1}{2}) - \\frac{1}{5})) + (\\frac{1}{5}) - \\frac{1}{8})) + (\\frac{1}{8}) - \\frac{1}{11})) + ....+ (\\frac{1}{299}) - \\frac{1}{302}))]

⟹ \\frac{1}{3}) [\\frac{1}{2}) - \\frac{1}{302})]

⟹ \\frac{1}{3}) [\\frac{151 - 1}{302})]

⟹ \\frac{1}{3}) [\\frac{150}{302})]

⟹ \\frac{50}{302}) = \\frac{25}{151})

The question is **"If a _{1} = \\frac{1}{2 × 5}) , a_{2} = \\frac{1}{5 × 8}) , a_{3} = \\frac{1}{8 × 11}),...., then a_{1} + a_{2} + a_{3} + ...... + a_{100} is" **

Choice A is the correct answer

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