Using Algebra Sequences Study Guide (page 2)
Introduction to Using Algebra Sequences
You can be moved to tears by numbers— provided they are encoded and decoded fast enough.
—Richard Dawkins (1941– ) British Scientist
In this lesson, you'll learn how to use algebra to find the nth term of arithmetic and geometric sequences, and how to find the exact values of terms within arithmetic and geometric sequences that contain algebraic terms.
A sequence is a set of numbers in which each number is generated according to a rule. We can use algebra to help us find the rule of a sequence or a certain term in a sequence. It might be easy to find the next term in a sequence once we have found the rule of the sequence, but what if we knew only a few numbers and wanted to find the hundredth term of the sequence? We wouldn't want to list all 100 terms!
An arithmetic sequence is a sequence in which the difference between any term and the term that precedes it is always the same. For instance, the sequence 3, 7, 11, 15, 19,…is an arithmetic sequence. We can find the rule of an arithmetic sequence by taking the difference between any two consecutive terms. In this sequence, the difference between consecutive terms is 4. The rule is +4, so the next term after 19 is 19 + 4, which is 23.
The numbers in a sequence are called terms. The first term of the sequence 3, 7, 11, 15, 19,…is 3, so we say that t1 = 3. The second term in the sequence is represented as t2 and the third as t3, and we say that the nth term is tn. The n represents an unknown place in the sequence. We use tn to make a general statement about a sequence.
We found the sixth term, t6, of the sequence by adding 4, the difference between any pair of consecutive terms, to the previous term, t5. If we know the first term in an arithmetic sequence and the difference between consecutive terms, we can find any term in the sequence using this formula: tn = t1 + (n – 1)d. In this formula, d represents the difference between any term and its previous term, and n represents the place of the term in the sequence.
We already know that the fifth term in the sequence is 19, but let's test the formula to be sure that it works. The first term, t1, is 3. The difference, d, is 4. We are looking for the fifth term, n = 5:
|tn = t1 + (n – 1)d|
|t5 = 3 + (5 – 1)4|
|t5 = 3 + (4)4|
|t5 = 3 + 16|
|t5 = 19|
What if the terms of a sequence are algebraic? Use two consecutive terms from the sequence and find their difference. Then, find the nth term of the sequence in terms of the variables used in the sequence.
|x + 2, 3x, 4x + 3, 6x + 1,…|
|We can find the fifth term, sixth term, or any other term using the formula tn = t1 + (n – 1)d. Let's find the seventh term.|
|First, find the difference between any pair of consecutive terms. Subtract the first term from the second term: 3x – (x + 2) = 2x – 2, so d = 2x – 2. The first term, t1, is x + 2. Because we are looking for the seventh term, n = 7:|
|t7 = (x + 2) + (7 – 1)(2x – 2)|
|t7 = (x + 2) + (6)(2x – 2)|
|t7 = (x + 2) + (12x – 12)|
|t7 = 13x – 10|
The seventh term of the sequence is 13x – 10. We can find the exact value of that term, because we know that the difference between any two consecutive terms in an arithmetic sequence is always the same. That's why it didn't matter which pair of consecutive terms we chose to subtract in order to find d. We found that the difference between the first term and the second term was 2x – 2. The difference between the second term and the third term is (4x + 3x) – 3x = x + 3. Because the difference between terms is always the same, 2x – 2 must equal x + 3. Set these two differences equal to each other and solve for x:
|2x – 2 = x + 3|
|x – 2 = 3|
|x = 5|
The value of x in this sequence is 5, which means that the seventh term, 13x – 10, is equal to 13(5) – 10 = 65 – 10 = 55.
When finding two different expressions that represent the difference, be sure to subtract terms in the order they appear in the sequence. The difference between the first two terms is the second term minus the first term, and the difference between the third term and the second term is the third term minus the second term.
A geometric sequence is a sequence in which the ratio between any term and the term that precedes it is always the same. For instance, the sequence 2, 4, 8, 16, 32,…is a geometric sequence. We can find the ratio, or rule, of a geometric sequence by dividing a term by the term that precedes it. In this sequence, the ratio between consecutive terms is 2. Because each term is equal to two times the previous term, the next term after 32 is 32(2), which is 64.
Just as we have a formula to find the nth term of an arithmetic sequence, we have a formula to find the nth term of a geometric sequence: tn =t1(rn – 1). In this formula, r represents the ratio of the sequence, and it is raised to the power n – 1. To find the eighth term of the sequence 2, 4, 8, 16, 32,…, we replace t1 with 2, n with 8, and r with 2:
|tn = t1(rn – 1)|
|t8 = 2(28 – 1)|
|t8 = 2(27)|
|t8 = 2(128)|
|t8 = 256|
|We can work with geometric sequences that have algebraic terms, too.|
When choosing two terms to use in order to find the ratio of a geometric sequence, try to pick two terms that, when divided, cause the variables to cancel out. If consecutive terms contain x with no constant added or subtracted, use these terms to find the ratio.
Find the sixth term of the sequence x – 4, 3x,15x, 75x,…
Begin by finding the ratio between consecutive terms. If we divide the third term by the second term, the variable
|tn = t1(rn – 1)|
|t6 = (t – 4)(56 – 1)|
|t6 = (x – 4)(55)|
|t6 = (x – 4)(3,125)|
|t6 = 3,125x – 12,500|
We know that the ratio between any two terms in the sequence is 5. That means that the second term, 3x, divided by the first term, x – 4, is equal to 5. Solve this equation for x:
|3x = 5x – 20|
|–2x = –20|
|x = 10|
Because x = 10, the sixth term is equal to 3,125(10) – 12,500 = 31,250 – 12,500 = 18,750.
It was helpful in finding the ratio that the variable x canceled out when we divided the third term by the second term. The fourth term divided by the third term is also 5. If the second term divided by the first term had been 5, we would have had no way of finding the value of x, because the ratios between the pairs of consecutive terms would have reduced to the same expression. If we are to find the value of the variable, we must always be given a sequence in which the ratio between two terms is a different algebraic expression from the ratios between the other pairs of terms. This is also true with arithmetic expressions: If the difference between every pair of consecutive terms is the same algebraic expression, then we cannot find the value of the variable.
Find practice problems and solutions for these concepts at Using Algebra: Sequences Practice Questions.
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