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If first three terms of a sequence 1 16 , a , b , 1 6 are in geometric series and the last three terms are in harmonic series, then the values of a and b will be

Options

  1. Aa = - 1 4 , b = 1
  2. Ba = - 1 12 , b = 1 9
  3. Cboth a and b are true
  4. DNeither a nor b

Correct answer

C. both a and b are true

Step-by-step solution

Given 1 16 , a and b are in GP. ⇒ a 2 = b 16 . . . i Also, a , b and 1 6 are in HP. ⇒ b = 2 × a × 1 6 a + 1 6 = 2 a 6 a + 1 . . . ii On solving equations i and ii , we get a = 1 12 , - 1 4 and b = 1 9 , 1 .

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