The sum of first n term is 3n2 - 4n
The sum of first n - 1 term is 3(n - 1)2 - 4(n - 1)
The difference of sum of (n – 1) and n will give us the value of an
The value of an = 6n - 7
Putting the value of n = 1, 2 to determine the first value of the series and the common difference
a1 = 6 x 1 - 7; a1 = -1
a2 = 6 x 2 - 7; a2 = 5
d = a2 - a1
= 5 - (-1)
= 6
Now to find the 12th term of series we use the formula of Arithmetic Progression:
a12 = a1 + (12 - 1)d
a12 = -1 + (11)6
a12 = -1 + 66
a12 = 65
Therefore, the value of 12th term of series is 65, whereas the value of nth term of series is 6n - 7.