Question

If $a_1,a_2,a_3,..$are in a harmonic progression with $a_1=5$ and $a_{20}=25$. The least positive integer $n$ for which $a_n<0$is

Answer 1

Step 1. Finding The least positive integer $\mathit{n}$ for which ${\mathit{a}}_{\mathbf{n}}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{}$

Given $a_1,a_2,a_3,...$are in HP

$\Rightarrow$ $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3}$are in AP

And $a_1=5$ and $a_{20}=25$

$\Rightarrow$$\frac{1}{a_1}=\frac{1}{5}$ and $\frac{1}{a_{20}}=\frac{1}{25}$

Let $d$ be the common difference,

Step 2. Find the common difference:

As we know,

$a_n=a+(n-1)d$

$\Rightarrow$ $\frac{1}{a_{20}}=\frac{1}{a_1}+19d$

$\Rightarrow$ $\frac{1}{a_{20}}–\frac{1}{a_1}=19d$

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

$\Rightarrow$ $-\frac{4}{25}=19d$

$\Rightarrow$ $d=-\frac{4}{25}\times \frac{1}{19}$

$\because a_n<0$

$a+(n-1)d<0$

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

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

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

$\Rightarrow$ $n>19\times \frac{5}{4}+1$

$\therefore n=25$

Hence, The least positive integer $n$ for which $a_n<0$is $\mathbf{25}$