Question

Find the number of terms in the series $20,19\frac{1}{3},18\frac{2}{3}$,__________ of which the sum is $300$, explain the double answer.

Answer 1

$a=20,d=-\frac{2}{3}$ and let $S_n=300$.
$\therefore \left(n/2\right)[2.20+(n-1\left)\right(-2/3\left)\right]=300$.
Simplifying $n^2-61n+900=0$
or $(n-25)(n-36)=0$
$n=25$ or $36$.
Since common ratio is negative and $S_{25}=S_{36}=300$, it shows that the sum of last eleven terms i.e., $T_{26},T_{27},....T_{36}$ is zero.