How to Solve Geometric Sequences? (+FREE Worksheet!)
Geometric sequences are number patterns where each term is found by multiplying the previous term by the same fixed number, called the common ratio. They appear in exponential growth and decay, financial calculations, and many real-world models. Understanding geometric sequences gives you a powerful tool for analyzing patterns and predicting future values in Algebra 1 and beyond.
Type the problems from this lesson and check each step.
Solve Geometric Sequences: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Compare termsLook for a common difference or common ratio.
- Choose term or sumDecide whether the question asks for one term or a total.
- Track nMake sure n is the position or number of terms the question uses.
Worked examples
Arithmetic sequence
- Each term adds 4.
- The common difference is 4.
- Add 4 to continue.
Geometric sequence
- Each term multiplies by 2.
- The common ratio is 2.
- Multiply 24 by 2.
Try one before moving on
Solve Geometric Sequences: pop-up practice
What Is a Geometric Sequence?
A geometric sequence is a list of numbers in which each term after the first is obtained by multiplying the previous term by a constant called the common ratio (\(\color{blue}{r}\)). If \(\color{blue}{r > 1}\), the sequence grows; if \(\color{blue}{0 < r < 1}\), the sequence shrinks; if \(\color{blue}{r < 0}\), the terms alternate in sign.
Example: \(\color{blue}{3, 6, 12, 24, 48, \ldots}\) The common ratio is \(\color{blue}{r = 2}\).
Key Formulas for Geometric Sequences
Finding the Common Ratio
Divide any term by the term before it:
\(\color{blue}{r = \frac{a_{n}}{a_{n-1}}}\)
Example: in \(\color{blue}{81, 27, 9, 3, \ldots}\) the ratio is \(\color{blue}{r = \frac{27}{81} = \frac{1}{3}}\).
Explicit (nth-Term) Formula
The nth term of a geometric sequence is:
an = a1 · r\(\color{blue}{n-1}\)
where \(\color{blue}{a_{1}}\) is the first term and \(\color{blue}{r}\) is the common ratio.
Recursive Formula
Each term is the previous term multiplied by \(\color{blue}{r}\):
\(\color{blue}{a_{n} = a_{n-1} \cdot r}\), with \(\color{blue}{a_{1}}\) given.
Step-by-Step Summary
- Identify the first term \(\color{blue}{a_{1}}\).
- Find the common ratio: \(\color{blue}{r = \frac{a_{2}}{a_{1}}}\) (verify with other consecutive pairs).
- Write the explicit formula: \(\color{blue}{a_{n} = a_{1} \cdot r^{n-1}}\).
- Substitute \(\color{blue}{n}\) to find any desired term.
Watch: Introduction to Geometric Sequences (Video Lesson)
Khan Academy introduces geometric sequences, common ratios, and the intuition behind the formula:
Geometric Sequences – Worked Examples
Example 1: Find the 5th term of the sequence \(\color{blue}{3, 6, 12, 24, \ldots}\)
\(\color{blue}{a_{1} = 3}\), \(\color{blue}{r = 2}\).
\(\color{blue}{a_{5} = 3 \cdot 2^{4} = 3 \cdot 16 = 48}\)
Example 2: Find the 5th term of \(\color{blue}{81, 27, 9, 3, \ldots}\)
\(\color{blue}{a_{1} = 81}\), \(\color{blue}{r = \frac{1}{3}}\).
\(\color{blue}{a_{5} = 81 \cdot (\frac{1}{3})^{4} = 81 \cdot \frac{1}{81} = 1}\)
Example 3: Find the 5th term of \(\color{blue}{2, -6, 18, -54, \ldots}\)
\(\color{blue}{a_{1} = 2}\), \(\color{blue}{r = -3}\).
\(\color{blue}{a_{5} = 2 \cdot (-3)^{4} = 2 \cdot 81 = 162}\)
Example 4: Find the common ratio when \(\color{blue}{a_{1} = 5}\) and \(\color{blue}{a_{4} = 40}\).
\(\color{blue}{40 = 5 \cdot r^{3} \Rightarrow r^{3} = 8 \Rightarrow r = 2}\)
More Practice: Using Explicit Formulas of Geometric Sequences (Video)
Khan Academy demonstrates how to apply the explicit formula to find specific terms and work backwards to find the common ratio:
Exercises for Geometric Sequences
- Find the common ratio: \(\color{blue}{5, 15, 45, 135, \ldots}\)
- Find the 6th term: \(\color{blue}{2, 6, 18, 54, \ldots}\)
- Find the 4th term: \(\color{blue}{256, 64, 16, \ldots}\)
- Write the explicit formula for: \(\color{blue}{4, 8, 16, 32, \ldots}\)
- Find \(\color{blue}{a_{7}}\) when \(\color{blue}{a_{1} = 1}\) and \(\color{blue}{r = -2}\).
Answers
- \(\color{blue}{r = 3}\)
- \(\color{blue}{a_{6} = 2 \cdot 3^{5} = 486}\)
- \(\color{blue}{a_{4} = 256 \cdot (\frac{1}{4})^{3} = 4}\)
- \(\color{blue}{a_{n} = 4 \cdot 2^{n-1}}\)
- \(\color{blue}{a_{7} = 1 \cdot (-2)^{6} = 64}\)
Frequently Asked Questions
What is a geometric sequence in math?
A geometric sequence is a list of numbers where each term is found by multiplying the previous term by the same fixed number, called the common ratio. For example, 3, 6, 12, 24, … multiplies by 2 each time, so the next term is 48. If the ratio is between 0 and 1, like in 80, 40, 20, 10, …, the terms shrink instead of grow.
What is the geometric sequence formula?
The explicit formula is aₙ = a₁ × rⁿ⁻¹, where a₁ is the first term, r is the common ratio, and n is the term number. For the sequence 5, 15, 45, …, a₁ = 5 and r = 3, so the 6th term is 5 × 3⁵ = 5 × 243 = 1,215. The recursive form is aₙ = aₙ₋₁ × r.
How do you find the common ratio of a geometric sequence?
Divide any term by the term right before it: r = a₂ ÷ a₁. In 2, −6, 18, −54, …, the ratio is −6 ÷ 2 = −3, and checking another pair (18 ÷ −6 = −3) confirms it. If the quotients between neighboring terms are not all the same, the sequence is not geometric. A negative ratio makes the signs alternate.
How do you find the sum of a geometric sequence?
Use Sₙ = a₁(1 − rⁿ) ÷ (1 − r), for any ratio r ≠ 1. For the first 5 terms of 4, 8, 16, 32, 64, a₁ = 4 and r = 2: S₅ = 4(1 − 32) ÷ (1 − 2) = 4(−31) ÷ (−1) = 124. Adding the terms directly, 4 + 8 + 16 + 32 + 64 = 124, confirms the answer.
Are geometric sequences exponential?
Yes. A geometric sequence is an exponential function whose inputs are limited to whole-number term positions. The formula aₙ = a₁ × rⁿ⁻¹ has the variable in the exponent, just like y = a × bˣ. For example, 2, 6, 18, 54, … matches the function f(n) = 2 × 3ⁿ⁻¹. Graphed, the points curve upward when r > 1 rather than lying on a straight line.
How do you find a geometric sequence with two terms?
Write both terms with the formula and divide to remove a₁. Suppose a₂ = 6 and a₅ = 162. Then a₁r⁴ ÷ a₁r = r³ = 162 ÷ 6 = 27, so r = 3. Next, a₁ = 6 ÷ 3 = 2, giving the sequence 2, 6, 18, 54, 162, … and the formula aₙ = 2 × 3ⁿ⁻¹.
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