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CAT Previous Year Paper | CAT Progressions Questions | Question 34

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Question 34 : If a1 = \\frac{1}{2 × 5}) , a2 = \\frac{1}{5 × 8}) , a3 = \\frac{1}{8 × 11}),...., then a1 + a2 + a3 + ...... + a100 is

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


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Explanatory Answer

Method of solving this CAT Question from Progressions

Given that a1 = \\frac{1}{2 × 5}) , a2 = \\frac{1}{5 × 8}) , a3 = \\frac{1}{8 × 11}),....,
We have to find a1 + a2 + a3 + ...... + a100
This can be done by partial fractions
an = \\frac{1}{(3𝑛 − 1)(3𝑛 + 2)})
a1 = 3 - 1 = 2 , a2 = 6 - 1 = 5 , a3 = 9 – 1 = 8.
100th term = \\frac{1}{((3 × 100) − 1)((3 × 100) + 2)})
a1 = \\frac{1}{2}) - \\frac{1}{5}) = \\frac{5 - 2}{2 × 5}) = \\frac{3}{2 × 5})
a2 = \\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 a1 = \\frac{1}{2 × 5}) , a2 = \\frac{1}{5 × 8}) , a3 = \\frac{1}{8 × 11}),...., then a1 + a2 + a3 + ...... + a100 is"

Hence, the answer is \\frac{25}{151})

Choice A is the correct answer

 

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