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)

25

Answer 2

PLAN nth term of HP, $t_n = \frac{1}{a+(n-1)n}$
Here, \hspace20mm a_1 = 5, a_{20} = 25 for HP
\therefore \hspace20mm \frac{1}{a}= 5, and $\frac{1}{a+19d} =25$
\Rightarrow \hspace20mm \frac{1}{5} = 19d =\frac{1}{25} \Rightarrow \, \, \, 19d = \frac{1}{25} - \frac{1}{5} = -\frac{4}{25}
\therefore \hspace20mm d =\frac{-4}{19 \times 25}
Since, \hspace15mm a_n < 0
\Rightarrow \hspace10mm \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 \hspace30mm n > 1 + \frac{95}{4} or $n > 24.75$
$\therefore$ Least positive value of n = 25