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Chapter 7: Problem 3

Evaluate the arithmetic series. $$ 302+305+308+\cdots+6002+6005+6008 $$

Short Answer

Expert verified
The sum of the arithmetic series \(302 + 305 + 308 + \cdots + 6002 + 6005 + 6008\) can be calculated using the formula \(S = \frac{n}{2}(first \; term + last \; term)\). With a common difference of 3, there are 1903 terms in the series. Thus, the sum is \(S = \frac{1903}{2}(302 + 6008) = 6,006,065\).

Step by step solution

01

Find the common difference

The common difference in an arithmetic series is the difference between consecutive terms. In this case, the common difference, labeled as 'd', is: \(d = 305 - 302 = 3\)
02

Find the number of terms

The total number of terms in the series (n) can be found using the formula: \(n = \frac{last \; term - first \; term}{common \; difference} + 1\) In our case, the last term is 6008, and the first term is 302: \(n = \frac{6008 - 302}{3} + 1\)
03

Simplify the number of terms

Now, we simplify the expression to find the number of terms: \(n = \frac{5706}{3} + 1\) \(n = 1902 + 1\) \(n = 1903\) So, the series has 1903 terms.
04

Find the sum of the arithmetic series

We can find the sum (S) of the arithmetic series using the formula: \(S = \frac{n}{2}(first \; term + last \; term)\) In our case, the first term is 302, the last term is 6008, and the total number of terms is 1903: \(S = \frac{1903}{2}(302 + 6008)\)
05

Simplify and calculate the sum

Now, we simplify the expression and then compute the sum: \(S = 951.5(6310)\) \(S = 6,006,065\) So, the sum of the given arithmetic series is 6,006,065.

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