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