Question

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 $a_n<0$ is

• A. 23
• B. 25
• C. 22
• D. 24

Answer 1

The correct option is B 25

As, $\frac{1}{a_{20}}=\frac{1}{25}=\frac{1}{5}+19d$

$\Rightarrow 19d=(\frac{1}{25}-\frac{1}{5})$

$\Rightarrow d=-\frac{5}{25\times 19}$

Also, $\frac{1}{a_n}=\frac{1}{5}-\frac{4}{25\times 19}(n-1)<0$

$\Rightarrow \frac{4(n-1)}{25\times 19}>\frac{1}{5}$

$\Rightarrow n>\frac{99}{4}$

and therefore; The least integral value of $n$ is $25$