The sum of the first \( n \) terms of an arithmetic sequence is given by \( S_n = 3n^2 + 5n \). Find the 10th term.
["Understanding Arithmetic Sequences: Finding the 10th Term Using ( S_n = 3n^2 + 5n )", "Arithmetic sequences are foundational in algebra and series summation, playing a key role in many mathematical and real-world applications. One common problem involves determining a specific term given the sum of the first ( n ) terms. In this article, we explore how to derive the 10th term of an arithmetic sequence when the sum of the first ( n ) terms is given by the formula:", "[\nS_n = 3n^2 + 5n\n]", "## The Sum of an Arithmetic Sequence", "For any arithmetic sequence, the sum of the first ( n ) terms, ( S_n ), can be expressed using the standard formula:", "[\nS_n = \frac{n}{2} \left(2a + (n-1)d\right)\n]", "where:\n- ( a ) is the first term,\n- ( d ) is the common difference.", "However, in this problem, we’re given ( S_n = 3n^2 + 5n ), a quadratic expression that simplifies finding sequence parameters.", "## Deriving the nth Term from the Sum", "To find the ( n )th term, ( a_n ), we use the relationship between the sum and the individual terms:", "[\na_n = S_n - S_{n-1}\n]", "Compute ( S_{n-1} ) by substituting ( n-1 ) into the given sum formula:", "[\nS_{n-1} = 3(n-1)^2 + 5(n-1)\n]\n[\n= 3(n^2 - 2n + 1) + 5n - 5\n]\n[\n= 3n^2 - 6n + 3 + 5n - 5\n]\n[\n= 3n^2 - n - 2\n]", "Now, subtract to find the ( n )th term:", "[\na_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2)\n]\n[\n= 3n^2 + 5n - 3n^2 + n + 2\n]\n[\n= 6n + 2\n]", "Thus, the general term of the sequence is:", "[\na_n = 6n + 2\n]", "## Finding the 10th Term", "Plugging ( n = 10 ) into the formula for ( a_n ):", "[\na_{10} = 6(10) + 2 = 60 + 2 = 62\n]", "## Conclusion", "Given the sum of the first ( n ) terms as ( S_n = 3n^2 + 5n ), we deduced that the ( n )th term is ( a_n = 6n + 2 ). Therefore, the 10th term is:", "[\n\boxed{62}\n]", "This approach efficiently leverages the connection between series sums and individual terms, offering a powerful tool for solving similar problems in algebra and mathematical competitions."]
