Question

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

Answer 1

Consider the sequence, $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots$
The above sequence is an AP with initial term a=20 and $d=19\frac{1}{4}-20=-\frac{3}{4}$
The general term, $t_n$, of the above sequence is
$t_n=a+(n-1)d$
$\Rightarrow t_n=20+(n-1)\left(\frac{-3}{4}\right)$ $[\because a=20andd=\frac{-3}{4}]$
$\Rightarrow t_n=20-(n-1)\left(\frac{3}{4}\right)$
To get the first negative term,
$t_n<0$
$\Rightarrow 20-(n-1)\left(\frac{3}{4}\right)<0\Rightarrow \frac{4\times 20-3(n-1)}{4}<0\Rightarrow 4\times 20-3(n-1)<0\Rightarrow 80-3n+3<0\Rightarrow 80+3<3n\Rightarrow 83<3n$
Number of terms should be in natural numbers.
Thus, the 28th term will be the first negative term of the above sequence
we know that,
$t_n=a+(n-1)d\Rightarrow t_{28}=20+(28-1)\left(\frac{-3}{4}\right)[Substitutingn=27,a=20andd=\frac{-3}{4}]\Rightarrow t_{28}=\frac{80+27\times (-3)}{4}\Rightarrow t_{28}=\frac{-1}{4}$