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Example: Determining if a Sequence Is Geometric

Determine whether each sequence is geometric and, if so, state the common ratio.

ⓐ 4, 8, 16, 32, 64, 128, dots

Divide consecutive terms: 84=2\frac{8}{4} = 2, 168=2\frac{16}{8} = 2, 3216=2\frac{32}{16} = 2, 6432=2\frac{64}{32} = 2, 12864=2\frac{128}{64} = 2. Every ratio equals 22, so the sequence is geometric with r=2r = 2.

ⓑ −2,6,−12,36,−72,216,…-2, 6, -12, 36, -72, 216, \dots

Divide consecutive terms: 6−2=−3\frac{6}{-2} = -3, −126=−2\frac{-12}{6} = -2, 36−12=−3\frac{36}{-12} = -3, −7236=−2\frac{-72}{36} = -2, 216−72=−3\frac{216}{-72} = -3. The ratios alternate between −3-3 and −2-2, so the sequence is not geometric and has no common ratio.

ⓒ 27,9,3,1,13,19,…27, 9, 3, 1, \frac{1}{3}, \frac{1}{9}, \dots

Divide consecutive terms: 927=13\frac{9}{27} = \frac{1}{3}, 39=13\frac{3}{9} = \frac{1}{3}, 13=13\frac{1}{3} = \frac{1}{3}, 131=13\frac{\frac{1}{3}}{1} = \frac{1}{3}, 1913=13\frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}. Every ratio equals 13\frac{1}{3}, so the sequence is geometric with r=13r = \frac{1}{3}.

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Updated 2026-06-17

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Intermediate Algebra @ OpenStax

Ch.12 Sequences, Series and Binomial Theorem - Intermediate Algebra @ OpenStax

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