Question

Which term of the sequence $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4}$ is the $1st$ negative term. Write the value of $n$ for that term.

Answer 1

Here $1^{st}$ term (a) $=20$ common difference
$d=19\frac{1}{4}-20=-\frac{3}{4}$
Let $n^{th}$ term of the given A.P be $1^{st}$ negative term
$\therefore a_n<0$
i.e $a+(n-1)d<0$
$\Rightarrow 20+(n-1)(-\frac{3}{4})<0$
$\Rightarrow \frac{83}{4}-\frac{3n}{4}<0$
$\Rightarrow 3n>83$
$\Rightarrow n>\frac{83}{3}$
$\Rightarrow n>27\frac{2}{3}$
Since, $28$ is the natural number just greater then $27\frac{2}{3}$
$\therefore 1^{st}$ negative term ${28}^{th}$.