The sum of the first n terms of an AP is given by Sn = (3n2 – 4n) .

Solution:

Given:

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

Now,

Sum of one term $=S_{1}=3(1)^{2}-4(1)$

$=3-4$

$=-1$

$\Rightarrow a_{1}=a=-1 \quad \ldots$ (1)

Hence, the first term is $-1$.

Sum of two term $s=S_{2}=3(2)^{2}-4(2)$

$=12-8$

$=4$

$\Rightarrow a_{1}+a_{2}=4 \quad \ldots(2)$

Subtracting (1) from (2), we get

$a_{2}=5$

$\Rightarrow a+d=5$

$\Rightarrow d=6 \quad(\because a=-1)$

Hence, the common difference is 6 .

Therefore,

$a_{n}=a+(n-1) d$

$=-1+(n-1)(6)$

$=-1+6 n-6$

$=6 n-7$

Hence, the $n^{\text {th }}$ term is $(6 n-7)$.

Hence,
(i) nth term is (6n – 7)
(ii) first term is –1
(iii) common difference is 6