The Arithmetic Sequence Formula

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Welcome back. Today I would like to conclude our discussion on arithmetic sequences that we began a few days ago. I recommend that you reread the first two posts on this topic before you move on to this one. Here are the links:

Arithmetic Sequences – Part One

Arithmetic Sequences – Part Two

Recall that an arithmetic sequence is a sequence of numbers such that the difference  between consecutive terms is constant. The number d is called the common difference of the arithmetic sequence. In this post I would like to give a very formal method for solving problems involving arithmetic sequences. Note that it is not necessary to learn this material in order to answer these types of questions on the ACT or SAT math subject tests, but every now and then the method given here gives a quicker way to solve a problem.

The arithmetic sequence formula

More advanced students may already know the arithmetic sequence formula:

a_n=a_1+(n-1)d

In this formula an is the nth term of the sequence. For example, a1 is the first term of the sequence.

Example 1

In the arithmetic sequence 20, 15, 10, 5, 0, -5, -10,… we have a= 20 and d = -5. Therefore we have

a_n=20+(n-1)(-5)=20-5n+5=25-5n

So for example, the 50th term of this arithmetic sequence is

a_{50}=25-5(50)=-225

Let’s revisit the second example from last week, but this time we will give a solution using the arithmetic sequence formula.

Example 2

Each term of a certain sequence is greater than the term before it. The difference between any two consecutive terms in the sequence is always the same number. If the fifth and ninth terms of the sequence are 33 and 97, respectively, what is the twelfth term?

Solution: Substituting 5 in for n and 33 in for an into the arithmetic sequence formula gives us

33 = a1 + 4d.

Similarly, substituting 9 in for n and 97 in for an into the arithmetic sequence formula gives us

 97 = a1 + 8d.

So we solve the following system of equations to find d.

97 = a1 + 8d
33 = a1 + 4d
64 =        4d

The last equation comes from subtraction. We now divide each side of this last equation by 4 to get

d = 16.

Finally, we add 16 to 97 three times to get 97 + 16(3) = 145.

Note that this solution is definitely more tedious than the previous solution from last week’s post. I usually prefer identifying arithmetic sequences with linear equations.

More Problems Involving Arithmetic Sequences

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