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JEE Main202229 Jul 2022Morning ShiftMathematicsSequences and SeriesActual

If 1 20 - a 40 - a + 1 40 - a 60 - a + … … + 1 180 - a 200 - a = 1 256 , then the maximum value of a is

Options

  1. A198
  2. B202
  3. C212
  4. D218

Correct answer

C. 212

Step-by-step solution

Given, 1 20 - a 40 - a + 1 40 - a 60 - a + … … + . . . +   1 180 - a 200 - a = 1 256 So, on simplifying we get, ⇒ 1 20 1 20 - a - 1 40 - a + 1 40 - a - 1 60 - a . . . . . + … + 1 180 - a - 1 200 - a = 1 256 ⇒ 1 20 1 20 - a - 1 200 - a = 1 256 ⇒ 20 - a 200 - a = 256 × 9 ⇒ a 2 - 220 a + 1696 = 0 ⇒ a = 8 , 212 So, maximum value of a is 212 .

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