Find the $$12^{th}$$ term of the progression $$S_n$$, where $$S_n = 1 + 3 + 7 + 13 + 21 + 31 + 43....$$

Observing the series shows that the difference between consecutive terms is in and AP. For instance: 3-1=2, 7-3=4, 13-7=6 and so on.

Hence, the general term (when the difference of terms is in an AP) is represented by a$$n^2$$+bn+c, where n is the number of that specific term in the series.

Putting n=1, we get a+b+c=1

Putting n=2, we get 4a+2b+c=3

Putting n=3, we get 9a+3b+c=7

On solving the three equations, a=1, b=(-1), and c=1

To find the $$12^{th}$$ term, we shall put n=12 and substituting  the values of a,b, and c in the equation a$$n^2$$+bn+c, we get it as 144-12+1= 133