Let a_1,a_2,a_3,..... be in harmonic progression with a_1 = 5 and a_20 = 25. The least positive integer n for which a_n \lt 0 is

Sequences and Series PYQ — JEE Advanced 2012 · Shift 2 · Q34 with Solution

View written solutionFree

Correct answer: D

If $a_1,a_2,a_3,\dots$ are in harmonic progression (H.P.), then their reciprocals are in arithmetic progression (A.P.).

So define

\frac{1}{a_n}

as an A.P.

Given: $a_1=5,\qquad a_{20}=25$ Thus, $\frac{1}{a_1}=\frac{1}{5},\qquad \frac{1}{a_{20}}=\frac{1}{25}$

Let

\frac{1}{a_n}=A+(n-1)d

where

Using $n=20$ :

\frac{1}{25}=\frac{1}{5}+19d

So,

19d=\frac{1}{25}-\frac{1}{5} = \frac{1-5}{25}=-\frac{4}{25}

Hence,

d=-\frac{4}{475}

Therefore,

\frac{1}{a_n}=\frac{1}{5}-(n-1)\frac{4}{475}

Since $a_n=\frac{1}{\frac{1}{a_n}},$ $a_n<0$ exactly when $\frac{1}{a_n}<0.$

So we need

\frac{1}{5}-(n-1)\frac{4}{475}<0

Multiply by $475$ :

95-4(n-1)<0

95-4n+4<0

99-4n<0

4n>99

n>24.75

Thus the least positive integer satisfying this is

n=25

For $n=24$ :

\frac{1}{a_{24}}=\frac{1}{5}-23\cdot\frac{4}{475}= \frac{95-92}{475}=\frac{3}{475}>0

so $a_{24}>0$ .

For $n=25$ :

\frac{1}{a_{25}}=\frac{1}{5}-24\cdot\frac{4}{475}= \frac{95-96}{475}=-\frac{1}{475}<0

so $a_{25}<0$ .

Hence the least such integer is indeed $25$ .

$25$ corresponds to Option D.