This is the basic and conceptual question from arithmetic progression chapter in which we are to calculate the value of n and also we need to find 21st term of this AP, if the sum of first m terms of an AP is (4m^2-m) and its nth term is 107.

RS Aggarwal, Class 10, Chapter 5C, question no 37

From the question it is given that,

an=107

The sum of first m terms of an A.P.Sm=4m^2−m

nth term Sn is =4n2−n

Then,

Sn−1=4(n−1)^2−(n−1)

=4(n^2−2n+1)−n+1

=4n^2−8n+4−n+1

=4n^2−9n+5

Therefore an=Sn−Sn−1107=4n^2−n−4n2+9n−5

107=8n−5

107+5=8n

112=8n

n=112/8

n=14

an=8n−5

a21=(8×14)−5

a21=168−5

a21=163