Question

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

• A. 39
• B. 40
• C. 41
• D. 42

Hint:

For AP sums, use $a_n = S_n - S_{n-1}$ to find the $n$-th term. Verify with the AP formula $a_n = a_1 + (n-1)d$.

Answer 1

The Correct Option is A

- Step 1: Understand the sum formula. Given $S_n = 2n^2 + n$, the $n$-th term is $a_n = S_n - S_{n-1}$.
- Step 2: Compute $S_{n-1$.} Substitute $n-1$:
\[ S_{n-1} = 2(n-1)^2 + (n-1) = 2(n^2 - 2n + 1) + n - 1 = 2n^2 - 4n + 2 + n - 1 = 2n^2 - 3n + 1. \] - Step 3: Find $n$-th term. $a_n = S_n - S_{n-1} = (2n^2 + n) - (2n^2 - 3n + 1) = 2n^2 + n - 2n^2 + 3n - 1 = 4n - 1$.
- Step 4: Calculate 10th term. For $n = 10$: $a_{10} = 4 \cdot 10 - 1 = 40 - 1 = 39$.
- Step 5: Verify with AP properties. First term ($n=1$): $a_1 = 4 \cdot 1 - 1 = 3$. Second term: $a_2 = 4 \cdot 2 - 1 = 7$. Common difference: $d = 7 - 3 = 4$. General term: $a_n = a_1 + (n-1)d = 3 + (n-1) \cdot 4 = 4n - 1$. For $n = 10$: $a_{10} = 3 + 9 \cdot 4 = 39$.
- Step 6: Check sum. $S_{10} = 2 \cdot 10^2 + 10 = 200 + 10 = 210$. AP sum: $S_n = \frac{n}{2} (a_1 + a_n)$. For $n = 10$, $a_1 = 3$, $a_{10} = 39$: $S_{10} = \frac{10}{2} (3 + 39) = 5 \cdot 42 = 210$. Matches.
- Step 7: Conclusion. The 10th term is 39, so option (1) is correct.