Question

find the nth term is an A.P., whose sum of n terms is 4n^2-n​

Answer 1

Answer :-

$a_n = 8n -4$

Given:-

$S_n = 4n^2-n$

To find:-

n th term in A. P

Solution:-

Replace n by (n - 1) ,

$S_n = 4 (n -1)^2 -n$

$S_n = 4(n^2 +1 -2n ) -n$

$S_n = 4n^2 + 4 -8n -n$

$S_n = 4n^2 -9n + 4$

Now, by using the formula for An :-

$\boxed{\sf{a_n = S_n - S_(n-1) }}$

$a_n = 4n^2 -n -(4n^2 -9n +4)$

$a_n = 4n^2 - n -4n^2 +9n -4$

$a_n = 4n^2 - 4n^2 + 9n -n -4$

$a_n = 8n -4$

hence, required nth term of an A. P is $a_n = 8n-4$