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 $n = 25$