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BITSAT 2022 Paper

Section: Mathematics

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Question : 113 of 130

Marks: +1, -0

Let

a_1, a_2, a_3 \ldots

be a harmonic progression with

a_1=5

a_{20}=25

. The least positive integer

n

a_n < 0

22
23
24
25

Solution: 👈: Video Solution

It is given that

a_1, a_2, a_3 \dots

are in

HP

.

Therefore,

\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots

are in

AP

Let

d

be the common difference of the

AP

.

\therefore \frac{1}{a_n} = \frac{1}{a_1} + (n-1)d \text{ and } \frac{1}{a_{20}} = \frac{1}{a_1} + 19d

\Rightarrow \frac{1}{a_n} = \frac{1}{5} + (n-1)d \text{ and } \frac{1}{25} = \frac{1}{5} + 19d

\Rightarrow \frac{1}{a_n} = \frac{1}{5} + (n-1)d \text{ and } d = \frac{-4}{19 \times 25}

\Rightarrow \frac{1}{a_n} = \frac{1}{5} - \frac{4(n-1)}{19 \times 25}

\Rightarrow \frac{1}{a_n} = \frac{95-4n+4}{19 \times 25}

\Rightarrow a_n = \frac{99-4n}{19 \times 25}

Now,

a_n < 0

\Rightarrow \frac{99-4n}{19 \times 25} < 0

\Rightarrow 99-4n < 0

\Rightarrow 4n > 99

\Rightarrow n > 24\frac{3}{4} \Rightarrow n \geq 25

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