Question

Which term of the sequence $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots$ will be the first negative term.

Answer 1

Let the required term be $n^{th}$ term .....
$\Rightarrow a_n<0$ (Given that it is -ve)
$a+(n-1)d<0$
$20+(n-1)\left(\frac{-3}{4}\right)<0$ $[19\frac{1}{4}-20=\frac{-3}{4}]$
$(n-1)\left(\frac{-3}{4}\right)<-20$
$(n-1)\left(\frac{3}{4}\right)>20$ (Multiplying by -1 on both sides)
$(n-1)>20\times \frac{4}{3}$
$n>\frac{80}{3}+1$
$n>\frac{83}{3}\Rightarrow n>27.66$
The integer just above 27.66 is 28
$\Rightarrow {28}^{th}term$ is the $1^{st}-v$
$a_{28}=20+27\left(\frac{3}{4}\right)$
$=\frac{80-81}{4}=-\frac{1}{4}$