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Chapter 7: Problem 4

Evaluate the arithmetic series. $$ 25+31+37+\cdots+601+607+613 $$

Short Answer

Expert verified
In the given arithmetic series, we find the common difference \(d = 6\), the number of terms \(n = 99\) and use the arithmetic series summation formula to calculate the sum: \(S_{99} = \frac{99}{2}(25 + 613)= 31,599\).

Step by step solution

01

Find the common difference

The difference between any two consecutive terms in the arithmetic series is the common difference. Let's find the difference between the first two terms and the last two terms: \(31 - 25 = 6\) \(613 - 607 = 6\) The common difference, d, between any two consecutive terms is 6.
02

Determine the number of terms

To find the number of terms in the arithmetic series, we will use the formula: \(n = \frac{(L - F)}{d} + 1\) Where n is the number of terms, L is the last term, F is the first term, and d is the common difference. In this case, L = 613, F = 25, and d = 6. Substitute the values and calculate the number of terms: \(n = \frac{(613 - 25)}{6} + 1\) \(n = \frac{588}{6} + 1\) \(n = 98 + 1\) \(n = 99\) There are a total of 99 terms in the arithmetic series.
03

Apply the arithmetic series summation formula

Now, let's use the arithmetic series summation formula to find the sum of the series: \(S_n = \frac{n}{2}(F + L)\) Where \(S_n\) is the sum of n terms, n is the number of terms, F is the first term, and L is the last term. In this case, n = 99, F = 25, and L = 613. Substitute the values and calculate the sum of the arithmetic series: \(S_{99} = \frac{99}{2}(25 + 613)\) \(S_{99} = \frac{99}{2}(638)\) \(S_{99} = 49.5 \cdot 638\) \(S_{99} = 31599\) So, the sum of the arithmetic series is 31,599.

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