Sequences and Series: Step-by-Step Worked Examples (8)

Question

The sum of the first $n$ terms of an AP is $S_n = 3n^2 + 5n$. Find the $n$-th term and the common difference.

Solution — Step by Step

Final answer: $T_n = 6n + 2$, $d = 6$.

Why This Works

For any sequence (not just AP), $T_n = S_n - S_{n-1}$ for $n \geq 2$, and $T_1 = S_1$. This identity recovers individual terms from cumulative sums.

For an AP, $T_n$ is linear in $n$ (form $a + (n-1)d$), so any sequence whose $T_n$ is linear in $n$ is automatically an AP. Here $T_n = 6n + 2$ is linear, common difference 6.

Alternative Method

For an AP, $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{d}{2}n^2 + (a - d/2)n$. Comparing with $S_n = 3n^2 + 5n$: $d/2 = 3 \Rightarrow d = 6$, and $a - d/2 = 5 \Rightarrow a = 8$. Then $T_n = a + (n-1)d = 8 + 6(n-1) = 6n + 2$.

Common Mistake

Computing $T_1$ from the formula $T_n = S_n - S_{n-1}$ which uses $S_0$ — undefined. Always handle $T_1 = S_1$ separately and apply the difference formula only for $n \geq 2$. Lucky for us, $T_1$ from the linear formula gives the same answer here, but that’s not guaranteed in general.