Suppose $S_n=2n+3n^2$. What is the $r$th term?

Question

For every $n$ the sum of the first $n$ terms of an arithmetic progression is $2n+3n^2$. What is the $r$th term of the sequence in terms of $r$.

Answer 1

A clever solution will be to notice the $rth$ term is $S_r-S_{r-1}$ using the equation...

$(3r^2+2r)-(3(r-1)^2+2(r-1))=6r-1$

Thus $$6r-1$$ is the desired equation.

Answer 2

The sum of first $n$ terms of series is $$S_n=2n+3n^2$$ The sum of first $n-1$ terms of series is $$S_{n-1}=2(n-1)+3(n-1)^2$$ $$S_{n-1}=3n^2-4n+1$$ Hence the $n$th term of the given series, $$T_n=S_n-S_{n-1}$$ $$=2n+3n^2-(3n^2-4n+1)$$
$$=6n-1$$ hence $r$th term of given series, $$T_r=6r-1$$ Above term shows that it is an Arithmetic Progression

Answer 3

Hint:

Use $S_n$ to find $S_1$ and $ S_2$.

We know that $S_1 = a_1$ and $ S_2 - S_1 = a_2$