Integers consecutive multiples

In this chapter, you find consecutive integers, consecutive even integers, consecutive multiples of fives, and so on. The word problems come in as puzzles to find the first, the middle, or the last in a list of consecutive integers. After an introduction on ways to find the sum of a large number of consecutive numbers, you ll see some interesting applications from seating charts to orchards. 

If n = 4 is what you got, and you want to find three consecutive multiples of 4, then your three integers are 4, 8, and 12. 

Pick a number, and you can make a list of integers that are multiples of that number. You can also make a symbolic list (using math symbols) of integers that are multiples of a number by starting with n and adding the number repeatedly. For example, a list of six consecutive multiples of 7, starting with 63 is 63, 70, 77, 84, 91, 98. Symbolically, a list of any six multiples of 7 is n, n + 7, n + 14, n + 21, n + 28, n + 35. 

Throughout the course of this book, we have looked at many word problems. Several problems involving distance and speed, percents, simple interest, and ratio and proportions have been reviewed. One other type of word problem not reviewed previously is consecutive integer problems. These problems are relatively easy to solve on multiple-choice tests. 

Consecutive integers are lists of integers that have a common difference between the terms. Odd and even numbers have a common difference of two between each term. Multiples of three have a common difference of three between the consecutive terms, and so on. 

The formula for finding the sum of a list of consecutive integers requires that you have the first and last terms in the list. The multiples of 4 are all four units apart. You have an arithmetic sequence with terms the difference of which is 4. (You can find more on arithmetic sequences in Setting the stage for the sums, earlier in this chapter.) So, to find the 20th term in the list of multiples of 4 that start with 60, use the formula an = + d(n - 1), giving you... 

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