Question

If $a_1,a_2,a_3,...$ are in 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

Answer: (D)

PLAN $n$ th 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 $n=25$