8.3 Arithmetic Sequences and Series

 

The following are examples of arithmetic ( air rith MEH tic ) sequences:

 

1, 2, 3, 4, 5, . . .

 

2, 5, 8, 11, 14, . . .

 

-2, 3, 8, 13, 18, . . .

 

What these sequences have in common is the fact that each term can be computed from the previous term by adding a constant which is fixed for that series.  This fixed term is called the common difference for the sequence.  In the first example, the common difference is 1, for the other two, it is 3 and 5, respectively.

 

The first sequence could be defined recursively as

 

c1 = 1,  cn+1 = 1 + cn

 

Exercise 8.3.1

 

Find recursive definitions for the other two arithmetic sequences above.

 

Solution

 

Consider the problem of finding the 87th term of the arithmetic sequence –31, -28, -25, -22, -19, . . .

 

The first term is –31 and the common difference is +3.  To find the 87th term it will be necessary to add the common difference to the first term a total of 86 times, since adding once gives the second term, adding twice gives the third term, and so on.  Thus, the 87th term would be –31 + 86(3) = 227.  This illustrates the principle that the nth term of an arithmetic sequence can be found by adding the common difference n – 1 times to the first term.  Stated as a formula:

 

 

Exercise 8.3.2

 

Find the 27th term of the arithmetic sequence 2, -1, -4, . . .

 

Solution

 

Exercise 8.3.3

 

Find the first term and the common difference of the arithmetic sequence given that c3 = 8 and c9 = 56.

 

Solution

 

There is a story told about the famous mathematician Gauss when he was a child.  Supposedly he and his classmates were challenged by their teacher to find the sum of all the integers from one to one hundred.  Gauss had the answer almost immediately.  When asked how he had done it he explained:  I added the first number to the last to get 1 + 100 =101, then I added the second number to the next to last to get 2 + 99 = 101.  I saw that I would continue to get a sum of 101 until I formed the last sum of 50 + 51 = 101.  Thus I would have 50 sums of 101 for a total of 50(101) = 5050.

 

What Gauss discovered is a method for finding the sum of an arithmetic sequence: Add the first term to the last term and multiply by half the number of terms.

 

Exercise 8.3.4

 

Use the method Gauss discovered to find the sum of the first 70 terms of the arithmetic sequence –11, -6, -1, 4, . . .

 

Solution

 

Exercise 8.3.5

 

Calculate the sum of the arithmetic series

 

 

Solution

 

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