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

Here first term, a=20.

and, $19\frac{1}{4}-20=\frac{77}{4}-20=\frac{77-80}{4}=\frac{-3}{4}$

$18\frac{1}{2}-19\frac{1}{4}=\frac{37}{2}-\frac{77}{4}=\frac{74-77}{4}=\frac{-3}{4}$

$17\frac{3}{4}-18\frac{1}{2}=\frac{71}{4}-\frac{37}{2}=\frac{71-74}{4}=\frac{-3}{4}$

$\therefore$ common difference, $d=\frac{-3}{4}$

Let $a_n$ be the first negative term

$\therefore a_n<0$

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

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

$\Rightarrow 80-3(n-1)<0$

$\Rightarrow 80-3n+3<0$

$\Rightarrow -3n+83<0$

$\Rightarrow -3n<-83$

$\Rightarrow 3n>83$

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

$\Rightarrow n>27.66$

Since n is a natural number, n=28.

$\Rightarrow n=28$

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

$=20+(28-1).\left(\frac{-3}{4}\right)=20+27.\left(\frac{-3}{4}\right)$

$=\frac{80-81}{4}$

$=-\frac{1}{4}$