Example

Example: Finding the Twelfth Term and General Term of a Geometric Sequence

Find the twelfth term of the sequence 3,6,12,24,48,96,3, 6, 12, 24, 48, 96, \dots and then write the general term.

First, identify the first term and the common ratio. The first term is a1=3a_1 = 3. To find the common ratio, divide each term by the one before it: 63=2\frac{6}{3} = 2, 126=2\frac{12}{6} = 2, 2412=2\frac{24}{12} = 2, 4824=2\frac{48}{24} = 2, 9648=2\frac{96}{48} = 2. Every ratio equals 22, so r=2r = 2.

To find the twelfth term, use the general term formula with a1=3a_1 = 3, r=2r = 2, and n=12n = 12:

an=a1rn1a_n = a_1 r^{n-1}

Substitute the values:

a12=32121a_{12} = 3 \cdot 2^{12-1}

Simplify the exponent:

a12=3211a_{12} = 3 \cdot 2^{11}

Evaluate:

a12=6,144a_{12} = 6{,}144

To write the general term, substitute a1=3a_1 = 3 and r=2r = 2 into the formula:

an=3(2)n1a_n = 3(2)^{n-1}

This example illustrates how to apply the general term formula when a sequence is given as a list of terms rather than with a1a_1 and rr stated directly. The first step is to determine both a1a_1 and rr from the listed terms by computing consecutive ratios, and then the formula is used both to find a specific term and to express the general rule for the sequence.

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Updated 2026-05-26

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Intermediate Algebra @ OpenStax

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

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