What is the formula for the nth term of a geometric sequence?

The nth term is a_n = a_1 · r^(n−1), where a_1 is the first term and r is the common ratio.

When does an infinite geometric series converge?

An infinite geometric series converges if and only if |r| < 1. Its sum is S = a_1 / (1 − r). If |r| ≥ 1 and a_1 ≠ 0, the series diverges.

What is the geometric mean of two numbers?

For three consecutive terms a, b, c of a geometric sequence, the middle term satisfies b² = ac, so b = ±√(ac). This is the geometric mean relationship.

Geometric Sequences and Series

Q: What is a geometric sequence?

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio r. For example, 1, 2, 4, 8, 16, … has r = 2.

Q: What is the geometric mean of two numbers?

For three consecutive terms a, b, c of a geometric sequence, the middle term satisfies b² = ac, so b = ±√(ac). This is the geometric mean relationship.

This tutorial explains geometric sequences and their sums. It covers the nth-term and sum formulas, the behavior of infinite geometric series, the geometric mean, and six worked examples with step-by-step solutions.

What Is a Geometric Sequence?

A geometric sequence is a sequence in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio, denoted \( r \).

Example 1: A Basic Geometric Sequence

The sequence \( 1, 2, 4, 8, 16, \ldots \) is geometric: each term is twice the previous one, so \( r = 2 \).

Fig.1 – The geometric sequence \(a_n = 2^{\,n-1}\) plotted on a linear vertical scale. The vertical gaps \(1, 2, 4, 8\) double at each step, producing the convex upward curve characteristic of exponential growth (contrast with the straight line of an arithmetic sequence).

Essential Formulas for Geometric Sequences

Recursive form:

\[ a_1 = \text{given}, \qquad a_n = r\,a_{n-1} \quad (n \ge 2) \]

Explicit (nth-term) formula:

\[ a_n = a_1\, r^{\,n-1} \]

Geometric mean: for three consecutive terms \(a\), \(b\), \(c\) of a geometric sequence,

\[ b^2 = ac \quad\Longrightarrow\quad b = \pm\sqrt{ac} \]

Sum of the first \(n\) terms (finite geometric series):

\[ S_n = a_1\,\frac{1 - r^n}{1 - r}, \quad r \neq 1 \]

Infinite geometric series (converges if \(|r| < 1\)):

\[ S_\infty = \frac{a_1}{1 - r} \]

Special case \(r=1\): every term equals \(a_1\), so \(S_n = n\,a_1\). (The formula above does not apply because the denominator would be zero.)

Worked Examples with Detailed Solutions

Example 2: Finding a Specific Term

Question: The 4th and 7th terms of a geometric sequence are \( \frac{1}{8} \) and \( \frac{1}{64} \) respectively. Find the 9th term.

View solution

With \(a_n = a_1 r^{n-1}\):

\[ a_4 = a_1 r^3 = \tfrac18, \qquad a_7 = a_1 r^6 = \tfrac{1}{64} \]

Divide \(a_7\) by \(a_4\) to eliminate \(a_1\):

\[ \frac{a_7}{a_4} = r^3 = \frac{1/64}{1/8} = \frac18 \]

Since the cube root is unique over the reals, \( r = \frac12 \).

Now use \(a_9 = a_7\,r^2\):

\[ a_9 = \frac{1}{64}\cdot\left(\frac12\right)^2 = \frac{1}{64}\cdot\frac{1}{4} = \frac{1}{256} \]

The 9th term is \( \dfrac{1}{256} \).

Example 3: Three Consecutive Terms Given Their Sum and Product

Question: Find three consecutive terms of a geometric sequence whose sum is \(234\) and whose product is \(157{,}464\).

View solution

Let the three terms be \(x\), \(xr\), and \(xr^2\) (with \(x\ne0\) and \(r\ne0\)).

Step 1 — Product:

\[ x\cdot xr\cdot xr^2 = x^3 r^3 = 157{,}464 \]

Since \(54^3 = 157{,}464\), the real cube root is unique:

\[ xr = 54 \quad\Longrightarrow\quad x = \frac{54}{r} \]

Step 2 — Sum:

\[ x + xr + xr^2 = 234 \quad\Longrightarrow\quad x(1 + r + r^2) = 234 \]

Substitute \(x = 54/r\):

\[ \frac{54}{r}(1 + r + r^2) = 234 \quad\Longrightarrow\quad 54r^2 - 180r + 54 = 0 \]

Divide by 18:

\[ 3r^2 - 10r + 3 = 0 \quad\Longrightarrow\quad (3r - 1)(r - 3) = 0 \]

So \(r = 3\) or \(r = \tfrac13\).

Step 3 — Find the terms:

If \(r = 3\): \(x = \frac{54}{3} = 18\). Terms: 18, 54, 162.
If \(r = \frac13\): \(x = \frac{54}{1/3} = 162\). Terms: 162, 54, 18.

Both orderings use the same three numbers, \(18, 54, 162\).

Example 4: Number of Terms for a Given Sum

Question: The first term of a geometric sequence is \(2000\) and the second is \(1000\). How many consecutive terms must be added to obtain a sum of \(3875\)?

View solution

The common ratio is \( r = \dfrac{a_2}{a_1} = \dfrac{1000}{2000} = \tfrac12 \).

Use \(S_n = a_1\dfrac{1 - r^n}{1 - r}\):

\[ 3875 = 2000\,\frac{1 - (1/2)^n}{1 - 1/2} = 4000\left(1 - \left(\tfrac12\right)^n\right) \]

\[ \frac{3875}{4000} = 1 - \left(\tfrac12\right)^n \quad\Longrightarrow\quad \frac{31}{32} = 1 - \left(\tfrac12\right)^n \]

\[ \left(\tfrac12\right)^n = \frac{1}{32} = \left(\tfrac12\right)^5 \quad\Longrightarrow\quad n = 5 \]

5 consecutive terms are needed.

Example 5: Proving a Sequence Is Not Geometric

Question: Prove that \(x\), \(x^2 + 1\), and \(x^3 + x\) cannot be three consecutive terms of a geometric sequence for any real \(x\).

View solution

If the three expressions were consecutive terms of a geometric sequence, the ratios of consecutive terms would be equal:

\[ \frac{x^2 + 1}{x} = \frac{x^3 + x}{x^2 + 1} \]

Cross-multiply:

\[ (x^2 + 1)^2 = x(x^3 + x) \]

Expand:

\[ x^4 + 2x^2 + 1 = x^4 + x^2 \]

Cancel \(x^4 + x^2\) from both sides:

\[ x^2 + 1 = 0 \quad\Longrightarrow\quad x^2 = -1 \]

Since \(x^2 \ge 0\) for every real \(x\), there is no real solution. Therefore the three expressions can never be consecutive terms of a real geometric sequence.

Example 6: Finding \(x\) from Algebraic Conditions

Question: The first three terms of a geometric sequence are \(x\), \(x^2 + 4\), and \(16x\). Find the values of these terms.

View solution

We need the ratios of consecutive terms to be equal:

\[ \frac{x^2 + 4}{x} = \frac{16x}{x^2 + 4} \]

Both denominators require \(x\ne0\) (since \(\frac{x^2+4}{x}\) is undefined at \(x=0\)) and \(x^2+4\ne0\) (automatic since \(x^2+4\ge4\)).

Cross-multiply:

\[ (x^2 + 4)^2 = 16x^2 \]

\[ x^4 + 8x^2 + 16 = 16x^2 \]

\[ x^4 - 8x^2 + 16 = 0 \quad\Longrightarrow\quad (x^2 - 4)^2 = 0 \]

\[ x^2 = 4 \quad\Longrightarrow\quad x = \pm 2 \]

Two valid term triples:

If \(x = 2\): terms are \(2,\ 2^2 + 4 = 8,\ 16(2) = 32\), with \(r = 4\).
If \(x = -2\): terms are \(-2,\ (-2)^2 + 4 = 8,\ 16(-2) = -32\), with \(r = -4\).

Infinite Geometric Series: Convergence and Divergence

When \(|r| < 1\), the terms shrink toward 0 and the sum converges:

\[ S_\infty = \frac{a_1}{1-r} \quad (|r| < 1) \]

Worked example: Find \( \displaystyle\sum_{n=1}^{\infty} 3\left(\frac{1}{2}\right)^{n-1} \).

View solution

Here \(a_1 = 3\) and \(r = \tfrac12\), with \(|r| < 1\), so the series converges:

\[ S_\infty = \frac{3}{1 - \frac12} = \frac{3}{\tfrac12} = 6 \]

The infinite sum equals 6.

If \(|r| \ge 1\) and \(a_1 \ne 0\), the terms do not shrink and the series diverges (the sum has no finite value). For example, \(1 + 2 + 4 + 8 + \cdots\) diverges.

Practice Problems

Try these on your own, then reveal each solution.

Problem 1

Find the 6th term of the geometric sequence \(3, 6, 12, \ldots\).

View solution

\(a_1 = 3\), \(r = 2\), so \(a_6 = 3\cdot 2^5 = 96\).

Problem 2

Find the sum of the first 8 terms of \(2, 6, 18, \ldots\).

View solution

\(a_1 = 2\), \(r = 3\), \(n = 8\):

\[ S_8 = 2\,\frac{1 - 3^8}{1 - 3} = 2\,\frac{1 - 6561}{-2} = 6560 \]

Problem 3

Find the sum of the infinite geometric series \( 8 + 4 + 2 + 1 + \cdots \).

View solution

\(a_1 = 8\), \(r = \tfrac12\), so \(S_\infty = \dfrac{8}{1 - 1/2} = 16\).

Related Practice

Ready for more? Explore additional math problems with solutions to sharpen your skills.