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BITSAT2022MathematicsSequences and SeriesActual

Let a₁, a₂, a₃ be a harmonic progression with a₁=5 and a₂₀=25 . The least positive integer n for which a_n < 0 , is

Options

  1. A22
  2. B23
  3. C24
  4. D25

Correct answer

D. 25

Step-by-step solution

It is given that a₁, a₂, a₃ . are in HP. Therefore, 1 a₁ , 1 a₂ , 1 a₃ , . . are in AP. Let d be the common difference of the AP. 1 a_n = 1 a₁ +(n-1) d and 1 a₂₀ = 1 a₁ +19 d 1 a_n = 1 5 +(n-1) d and 1 25 = 1 5 +19 d 1 a_n = 1 5 +(n-1) d and d= -4 19 25 1 a_n = 1 5 - 4(n-1) 19 25 1 a_n = 95-4 n+4 19 25 a_n= 99-4 n 19 25 Now, a_n 99 n>24 3 4 n 25

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