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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2022 · Mathematics · Sequences and Series

Let a_1, a_2, a_3 be a harmonic progression with a_1=5 and a_ 20...

Let a1,a2,a3…a_1, a_2, a_3 \ldots be a 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

It is given that a1,a2,a3…a_1, a_2, a_3 \ldots. are in HP. Therefore, 1a1,1a2,1a3,…….\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \ldots.. are in AP. Let dd be the common difference of the AP. ∴1an=1a1+(n−1)d\therefore \quad \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d and 1a20=1a1+19d⇒1an=15+(n−1)d\frac{1}{a_{20}}=\frac{1}{a_1}+19 d\Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}+(n-1) d and 125=15+19d⇒1an=15+(n−1)d\frac{1}{25}=\frac{1}{5}+19 d\Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}+(n-1) d and d=−419×25⇒1an=15−4(n−1)19×25⇒1an=95−4n+419×25⇒an=99−4n19×25d=\frac{-4}{19 \times 25}\Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}-\frac{4(n-1)}{19 \times 25}\Rightarrow \quad \frac{1}{a_n}=\frac{95-4 n+4}{19 \times 25}\Rightarrow \quad a_n=\frac{99-4 n}{19 \times 25} Now, an<0⇒99−4n19×25<0⇒99−4n<0⇒4n>99⇒n>2434⇒n≥25a_n < 0\Rightarrow \quad \frac{99-4 n}{19 \times 25} < 0\Rightarrow \quad 99-4 n < 0\Rightarrow \quad 4 n > 99\Rightarrow \quad n > 24 \frac{3}{4} \Rightarrow n \geq 25
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