Question

Find the term of the series $25,22\frac{3}{4},20\frac{1}{2},18\frac{1}{4}.....$ which is numerically smallest.

Answer 1

The given series is an A.P., i.e., $a=25,d=-9/4.$
$T_n=a+(n-1)d=(25+\frac{9}{4})-\frac{9}{4}n$
or $T-n=\frac{109}{4}-\frac{9}{4}n$
Now $T_n$ will be negative if $\frac{109}{4}-\frac{9}{4}n<0$ or $n>12\frac{1}{9}$.
Above shows that $T_{13}$ will be the first negative term and hence $T_{12}$ will be the smallest positive terms. $T_{13}=-2,T12=\frac{1}{4}$ is numerically smallest.