Using Algebra Sequences Study Guide (page 2)

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Updated on Oct 3, 2011

Geometric Sequences

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 will cancel out, because 15x ÷ 3x = 5. The ratio, r, is 5. The ?rst term in the sequence, t1, is x – 4. Because we are looking for the sixth term, n = 6:

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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