Question

The sum of the first $n$ terms of an arithmetic progression is $3n^2 + 2n$. What is the 10th term?

• A. 59
• B. 60
• C. 61
• D. 62

Hint:

For arithmetic progression, use $a_n = S_n - S_{n-1}$ to find the $n$-th term.

Answer 1

The Correct Option is C

- Step 1: The sum of the first $n$ terms is $S_n = 3n^2 + 2n$.
- Step 2: The $n$-th term is $a_n = S_n - S_{n-1}$.
- Step 3: Compute $S_{n-1} = 3(n-1)^2 + 2(n-1) = 3(n^2 - 2n + 1) + 2n - 2 = 3n^2 - 6n + 3 + 2n - 2 = 3n^2 - 4n + 1$.
- Step 4: So, $a_n = (3n^2 + 2n) - (3n^2 - 4n + 1) = 3n^2 + 2n - 3n^2 + 4n - 1 = 6n - 1$.
- Step 5: For the 10th term, $a_{10} = 6 \times 10 - 1 = 60 - 1 = 59$.
- Step 6: Check options: Option (1) is 59, but recompute for accuracy: $a_n = 6n - 1$ is correct, but verify $a_{10} = S_{10} - S_9$. $S_{10} = 3 \times 10^2 + 2 \times 10 = 320$, $S_9 = 3 \times 9^2 + 2 \times 9 = 261$, so $a_{10} = 320 - 261 = 59$. Option (3) seems incorrect; correct answer is (1).