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. 22
• B. 23
• C. 24
• D. 25

Answer 1

The correct option is D 25

$\because a_1,a_2,a_3,....$ are in H.P.
$\therefore \frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3}.......$ are in A.P.
$\therefore \frac{1}{a_1}=\frac{1}{5}and\frac{1}{a_{20}}=\frac{1}{25}$
$\frac{1}{a_1}+19d=\frac{1}{a_{20}}\Rightarrow \frac{1}{5}+19d=\frac{1}{25}\Rightarrow d=\frac{-4}{475}$
Now $\frac{1}{a_n}=\frac{1}{5}+(n-1)\left(\frac{-4}{475}\right)$
Clearly $a_n<0if\frac{1}{a_n}<0\Rightarrow \frac{1}{5}-\frac{4n}{475}+\frac{4}{475}<0$
$\Rightarrow -4n<-99orn>\frac{99}{4}=24\frac{3}{4}\therefore n\ge 25$
Hence least value of n is 25.