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8.3 Arithmetic Sequences and Series

What is an arithmetic sequence?

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

\(1,2,3,4,\cdots\)

\(2,5,8,11,\cdots\)

\(-2,3,8,13,\cdots\)

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 sequence. 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.

An arithmetic sequence is determined solely by its first term and its common difference. If one knows the first term and the common difference, then one can write the sequence.

For the first sequence above, the first term is \(c_1=1\) and the common difference is \(d=1\). Therefore

\(c_2=d+c_1=1+1=2\)

\(c_3=d+c_2=1+2=3\)

\(c_4=d+c_3=1+3=4\)

\(\vdots\)

\(c_{n+1}=d+c_n=1+c_n\)

The recursive formula for this sequence is \(c_1=1\quad c_{n+1}=1+c_n\)

In general, the recursive formula for an arithmetic sequence with first term \(c\) and common difference \(d\) is

\[c_1=c,\quad c_{n+1}=d+c_n\]

Exercise 8.3.1

Find recursive formulas for the other two arithmetic sequences above.

See Solution

The general formula for an arithmetic sequence

Consider the problem of finding the \(87\)th term of the arithmetic sequence \(31, -28, -25, -22, -19, \cdots\)

The first term is \(-31\) and the common difference is \(+3\). To find the \(87\)th 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 \(87\)th term would be \(–31 + 86(3) = 227\).

This illustrates the principle that the \(n\)-th term of an arithmetic sequence can be found by adding the common difference \(n – 1\) times to the first term.

Thus the general formula for an arithmetic sequence is

\[c_n=c_1+(n-1)d\]

Exercise 8.3.2

Find the \(27\)th term of the arithmetic sequence \(2, -1, -4, \cdots\)

See Solution

Exercise 8.3.3

Find the first term and the common difference of the arithmetic sequence given that \(c_3=8\) and \(c_9=56\).

See Solution

The sum of an arithmetic sequence: Arithmetic series

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.

\[s_n=\frac{n}{2}\left(c_1+c_n\right)\]

We can obtain an alternate version of this formula if we combine it with the formula for \(c_n\) from above:

\[s_n=\frac{n}{2}\left[2c_1+(n-1)d)\right]\]

Exercise 8.3.4

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

See Solution

Exercise 8.3.5

Calculate the sum of the arithmetic series:

\[\sum_{p=2}^{12}(2p-1)\]

See Solution