Problem 82 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use a graphing utility to find the partial sum. $$\sum_{j=1}^{200}(10.5+0.025 j)$$

Step-by-Step Solution

Answer

The partial sum of the series is 2597.5.

Step 1: Identify the Arithmetic Series Terms

Identify the arithmetic series components in the exercise. The series starts from the first term 10.5 (which is \(a_1\)), increments by 0.025 (the common difference, \(d\)), and ends at the 200th term (the number of terms, \(n\)). Thus, \(a_1 = 10.5\), \(d = 0.025\) and \(n = 200\).

Step 2: Apply the Arithmetic Series Formula

The sum of the arithmetic series is found by applying the formula: \(S = n/2 * [2a + (n - 1)d]\). Substituting the identified components into this formula gives: \(S = 200/2 * [2 * 10.5 + (200 - 1) * 0.025]\). This simplifies to \(S = 100 * [21 + 199 * 0.025]\).

Step 3: Calculate the Sum

Simplify the expression by carrying out the arithmetic operations: 199*0.025 equals to 4.975, resulting the equation \(S = 100 * [21 + 4.975]\). Further simplify this equation: \(S = 100 * 25.975\). Lastly, calculate the sum \(S = 2597.5\).