Question

If $a_1,a_2,a_3,\dots \dots$ are in a harmonic progression with $a_1=5$ and $a_{20}=25$, then, the least positive integer n for which $a_n<0$, is

• A. 22
• B. 23
• C. 24
• D. 25

Answer 1

The correct option is D

25

nth term of HP, $t_n=\frac{1}{a+(n-1)n}$
Here, $a_1=5,a_{20}=25$ for HP
$\therefore \frac{1}{a}=5$ and $\frac{1}{a+19d}=25$
$\Rightarrow \frac{1}{5}+19d=\frac{1}{25}$
$\Rightarrow 19d=\frac{1}{25}-\frac{1}{5}=-\frac{4}{25}$
$\therefore d=\frac{-4}{19\times 25}$
Since, $a_n<0$
$\Rightarrow \frac{1}{5}+(n-1)d<0$
$\Rightarrow \frac{1}{5}-\frac{4}{19\times 25}(n-1)<0\Rightarrow (n-1)>\frac{95}{4}$
$\Rightarrow n>1+\frac{95}{4}$ or $n>24.75$
$\therefore$ Least positive value of n = 25