Question

The sum of the first n terms of an A.P. is $3n^2-4n$. Find the nth term of this A.P.

Answer 1

Let the sum of first n terms of the A.P.=Sn

Given: $Sn=3n^2–4n$ ...(i)

Now,

Replacing n by (n –1) in (i), we get,

$Sn–1=3(n–1{)}^2–4(n–1)$
nth term of the A.P. $an=Sn–(Sn–1)$

$\therefore an=(3n^2–4n)–\left[3\right(n–1{)}^2–4(n–1)]$

$\Rightarrow an=3[n^2–(n–1{)}^2]–4[n–(n–1)]$

$\Rightarrow an=3(n^2–n^2+2n–1)–4(n–n+1)$

$\Rightarrow an=3(2n–1)–4$

$\Rightarrow an=6n–3–4$

$\Rightarrow an=6n–7$

Thus, the nth term of the $A.P=6n–7.$