Question

If sn, the sum of first n terms of an



a.P. Is given by sn = 3n2 - 4n then find its nth term.

Answer 1

Answer :- an = 6n - 7

Given:-

$S_n = 3n^{2} - 4n$

To find :-

It's nth term or $a_n$

Solution:-

We have ,

$S_n = 3n^{2} - 4n$

Replace n by n - 1 ,

we get,

$\implies$ $S_(n - 1) = 3 (n -1 ) ^{2} - 4 (n - 1)$

$\implies$ $S_(n-1) = 3 ( n^2 + 1 -2n ) - 4 ( n-1)$

$\implies$ $S_n - 1 = 3 n^2 + 3 -6n -4n +4$

$\implies$ $S_(n-1) = 3n^2 + 7 - 10 n$

Now use, the relation between Sn, S (n-1) and an.

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

Then,

$\implies$ $a_n = 3n^2 - 4n - ( 3n^2 + 7 -10n)$

$\implies$ $a_n = 3n^2 -4n -3n^2 -7 +10n$

$a_n = 6n - 7$