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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2012 · Paper 2 · Mathematics · Sequences and Series

Let a_ 1 , a_ 2 , a_ 3 , . . be in harmonic progression with a_ 1...

Let a1,a2,a3,…..a_{1}, a_{2}, a_{3}, \ldots . . be in harmonic progression with a1=5a_{1}=5 and a20=25a_{20}=25. The least positive integer nn for which an<0a_{n} < 0 is

Options

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

Correct answer

D. 25

Step-by-step solution

∵a1,a2,a3,……\because a_{1}, a_{2}, a_{3}, \ldots \ldots are in H.P. ∴1a1,1a2,1a3…… are in A.P. ∴1a1=15 and 1a20=1251a1+19d=1a20⇒15+19d=125⇒d=−4475 Now 1an=15+(n−1)(−4475) Clearly an<0, if 1an<0⇒15−4n475+4475<0⇒−4n<−99 or n>994=2434∴n≥25∴ Least value of n is 25.\begin{array}{l} \therefore \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} \ldots \ldots \text{ are in A.P. } \\ \therefore \quad \frac{1}{a_{1}}=\frac{1}{5} \text{ and } \frac{1}{a_{20}}=\frac{1}{25} \\ \frac{1}{a_{1}}+19 d=\frac{1}{a_{20}} \Rightarrow \frac{1}{5}+19 d=\frac{1}{25} \Rightarrow d=\frac{-4}{475} \\ \text{ Now } \frac{1}{a_{n}}=\frac{1}{5}+(n-1)\left(\frac{-4}{475}\right) \\ \text{ Clearly } a_{n} < 0, \text{ if } \frac{1}{a_{n}} < 0 \Rightarrow \frac{1}{5}-\frac{4 n}{475}+\frac{4}{475} < 0 \\ \Rightarrow \quad-4 n < -99 \text{ or } n>\frac{99}{4}=24 \frac{3}{4} \quad \therefore n \geq 25 \\ \therefore \quad \text{ Least value of } n \text{ is } 25 . \end{array}
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