Discussion Forum

Let $a_{1},a_{2},a_{3}.......$ be in a harmonic progression with  $a_{1}=5$ and $a_{20}=25$. The least  positive integer n for which $a_{n} <0$ is 

Answer:

Explanation:

 Concept involved $n^{th}$ term of HP 

  $t_{n}=\frac{1}{a+(n-1)d}$

 Sol. 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

 $\therefore$  $\frac{1}{5}+(n-1)d <0$

$\Rightarrow$  $\frac{1}{5}-\frac{4}{19 \times 25}(n-1) <0$

 $\Rightarrow$  $(n-1) > \frac{95}{4}$

 $\Rightarrow$  $n > 1+\frac{95}{4}$ or $n > 24.75$

 least positive  value of n=25