Determine \( n \) if: \( 4+13+22+\ldots .( \) to \( n \) terms \( )=539 \) mine \( m \) in each of the following arithmetic series:

To find \( n \) such that the sum of the series \( 4 + 13 + 22 + \ldots \) equals \( 539 \), we first note that the given series is an arithmetic sequence where the first term \( a = 4 \) and the common difference \( d = 9 \) (since \( 13 - 4 = 9 \)). The \( n \)-th term of an arithmetic sequence can be calculated using the formula: \[ a_n = a + (n-1)d = 4 + (n-1) \cdot 9 \] Thus, the \( n \)-th term can be simplified to: \[ a_n = 4 + 9n - 9 = 9n - 5 \] The sum \( S_n \) of the first \( n \) terms in an arithmetic sequence can be calculated using the formula: \[ S_n = \frac{n}{2} \times (a + a_n) = \frac{n}{2} \times (4 + (9n - 5)) = \frac{n}{2} \times (9n - 1) \] We set this equal to \( 539 \): \[ \frac{n}{2} \times (9n - 1) = 539 \] Multiplying both sides by 2 gives: \[ n(9n - 1) = 1078 \] This leads to the quadratic equation: \[ 9n^2 - n - 1078 = 0 \] Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 9 \), \( b = -1 \), and \( c = -1078 \): \[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 9 \cdot (-1078)}}{2 \cdot 9} \] \[ n = \frac{1 \pm \sqrt{1 + 38808}}{18} \] \[ n = \frac{1 \pm \sqrt{38809}}{18} \] \[ n = \frac{1 \pm 197}{18} \] Calculating the two potential solutions gives: 1. \( n = \frac{198}{18} = 11 \) 2. \( n = \frac{-196}{18} \) (not viable since \( n \) must be positive) Thus, the solution is: \[ n = 11 \] In context of other arithmetic series, it’s common to find patterns and transformations that simplify calculations. Recognizing that arithmetic series have a linear structure simplifies both calculations and understanding of how the terms behave as numbers increase. If you're considering exploring multiple arithmetic series, try varying the first term and the common difference, observing how those choices affect the sum and number of terms!