Find the term of the series `25 ,22 ,3/4,20 1/2,18 1/4` which is numerically the smallest.

Explanation

The given series is an A.P, with a=25,d=-9/4.
`T_(n)=a+(n-1)d=(25+9/4)-9/4n`
or `T_(n)=109/4-9/4n`
Now `T_(n)` will be negative if `109/4-9/4nlt0orngt121/9`
So, `T_(13)` will be the first negative term and hence `T_(12)` will be the smallest positive terms. `T_(13)=-2,T_(12)=1/4` is numerically smallest.