Question

Which term of the progression $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4}$..is the first negative term?

Answer 1

Given, A.P. is $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4}...$

= $20,\frac{77}{4},\frac{37}{2},\frac{71}{4},....$

Here, a = 20, d = $\frac{77}{4}-20=\frac{77-80}{4}=\frac{-3}{4}$

Let $a_n$ is first negative term

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

$\Rightarrow 20+(n-1)(-\frac{3}{4})<0$ $\Rightarrow 20+\frac{3}{4}<\frac{3}{4}n$

$\Rightarrow \frac{83}{4}<\frac{3}{4}n$

$\Rightarrow n>\frac{83}{4}\times \frac{4}{3}$

$\Rightarrow n>\frac{83}{3}=27.66$

${28}^{th}$ term will be the first negative term of given A.P.