Question

Find the least number of terms of the series $19+18\frac{1}{5}+17\frac{2}{5}+...$ whose sum is negative. Also calculate the exact sum.

Answer 1

Solution :

Series is $A.P$

$a=19d=\frac{-4}{5}$

Sum of $n$ terms $=\frac{n}{2}(38+(n-1\left)\frac{-4}{5}\right)$

$=19n\frac{4n^2}{2\times 5}+\frac{4n}{2\times 5}$

$=19n-\frac{2n^2}{5}+\frac{2n}{5}$

$=\frac{97n}{5}-\frac{2n^2}{5}$

$\frac{97n}{5}-\frac{2n^2}{5}<0.$

$n(97-2n)<0$

$n>0$

then $97-2n<0$

$\frac{97}{2}<n$

Least $n=49$ for negative sum of series.

Exact sum $=\frac{49(97-2\times 49)}{5}=\frac{-49}{5}$