Question

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

• A. 25
• B. 36
• C. 49
• D. both 25 and 36

Answer 1

The correct option is D both 25 and 36

For series $20,19\frac{1}{3},18\frac{2}{3}$
$a=20,d=-\frac{2}{3}$ and $S_n=300$
$\Rightarrow n^2-61n+900=0\Rightarrow [n-25][n-36]=0$
$\Rightarrow n=25$ or $36$
Hence $S_{25}=S_{36}=300$
But as $d$ is negative sum of last $11$ term i.e $T_{26},T_{27},...T_{36}$ is zero