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,

anâ€‹=107

The sum of firstÂ mÂ terms of anÂ A.P.Smâ€‹=4m^2âˆ’m

nthÂ Â termÂ Snâ€‹Â isÂ =4n2âˆ’n

Â Then,Â

Snâˆ’1â€‹=4(nâˆ’1)^2âˆ’(nâˆ’1)

Â Â Â Â =4(n^2âˆ’2n+1)âˆ’n+1

Â Â Â Â =4n^2âˆ’8n+4âˆ’n+1

Â Â Â Â =4n^2âˆ’9n+5

Therefore an=Snâ€‹âˆ’Snâˆ’1â€‹107=4n^2âˆ’nâˆ’4n2+9nâˆ’5

107=8nâˆ’5

107+5=8n

112=8n

n=112/8

n=14â€‹

anâ€‹=8nâˆ’5

a21â€‹=(8Ã—14)âˆ’5

a21â€‹=168âˆ’5

a21â€‹=163â€‹