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Chapter 7: Problem 10

Evaluate the arithmetic series. $$ \sum_{k=10}^{900}(3 k-2) $$

Short Answer

Expert verified
The sum of the arithmetic series \(\sum_{k=10}^{900}(3 k-2)\) is 1,214,363.

Step by step solution

01

Find the first term and the last term

To find the first term in the series, substitute the lower limit of the summation, which is \(k = 10\) into the given expression: $$ a_1 = 3(10) - 2 = 30 - 2 = 28 $$ Similarly, substitute the upper limit of the summation, which is \(k = 900\), into the given expression to find the last term: $$ a_n = 3(900) - 2 = 2700 - 2 = 2698 $$ Thus, the first term, \(a_1\), in the series is 28, and the last term, \(a_n\), is 2698.
02

Determine the number of terms in the series

Since the series starts at \(k = 10\) and ends at \(k = 900\), we can find the number of terms, \(n\), by calculating the difference between the upper and lower limits and then adding 1: $$ n = (900 - 10) + 1 = 890 + 1 = 891 $$ There are 891 terms in the series.
03

Calculate the sum of the series using the arithmetic series formula

Now that we have the first term, the last term, and the number of terms in the series, we can use the arithmetic series formula to find the sum, \(S\), of the series: $$ S = \frac{n(a_1 + a_n)}{2} $$ Plug in the values we found for the first term, the last term, and the number of terms: $$ S = \frac{891(28 + 2698)}{2} = \frac{891(2726)}{2} = 891 \cdot 1363 = 1214363 $$ Therefore, the sum of the arithmetic series is 1,214,363.

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