Question

Find the smallest positive integer $n$ such that $t_n$ of the arithmetic sequence $20,19\frac{1}{4},18\frac{1}{2},....$ is negative?

Answer 1

Here we have $a=20,d=19\frac{1}{4}-20=-\frac{3}{4}$.
We want to find the first positive integer $n$ such that $t_n<0$.
This is same as solving $a+(n-1)d<0$ for smallest $nϵN$.
That is solving $20+(n-1)(-\frac{3}{4})<0$ for smallest $nϵN$.
Now, $(n-1)(-\frac{3}{4})<-20$
$\Rightarrow (n-1)\times \frac{3}{4}>20$ (The inequality is reversed on multiplying both sides by $-1$)
$\Rightarrow n-1>20\times \frac{4}{3}=\frac{80}{3}=26\frac{2}{3}$.
This implies $n>26\frac{2}{3}+1$. That is, $n>27\frac{2}{3}=27.66$
Thus, the smallest positive integer $nϵN$ satisfying the inequality is $n=28$.
Hence, the ${28}^{th}$ term, $t_{28}$ is the first negative term of the A.P.