Geometric Sequence Formulas: nth Term, Sum, and Examples

Geometric sequence formulas are the equations used to find any term in a geometric sequence and to compute the sum of its terms. There are three core formulas: the nth term formula, the sum of the first n terms, and the sum of an infinite geometric sequence. A geometric sequence is one where each term is the previous term multiplied by a fixed number called the common ratio.

Quick Reference: All Geometric Sequence Formulas

In every formula, a is the first term, r is the common ratio, and n is the number of terms.

nth Term Formula of a Geometric Sequence

The nth term formula:

aₙ = a · r^(n−1)

Variable Key

When to Use It

Use the nth term formula when the first term and common ratio are known, and a specific term is needed without listing every term before it. The exponent is (n − 1), not n — it counts how many times r has been multiplied to reach position n.

Some textbooks index sequences starting from a₀ instead of a₁. In that case, the formula becomes aₙ = a · rⁿ. Check the textbook's convention before applying the formula.

Worked Example

Find the 7th term of the sequence 3, 6, 12, 24, …

a = 3, r = 2, n = 7

a₇ = 3 · 2^(7−1) = 3 · 2⁶ = 3 · 64 = 192

Answer: a₇ = 192

Sum of the First n Terms Formula

The sum formulas:

Sₙ = a(1 − rⁿ)/(1 − r), when r ≠ 1

Sₙ = a(rⁿ − 1)/(r − 1), when r ≠ 1

Sₙ = na, when r = 1

The Two Forms Are the Same Formula

The first two forms produce identical answers. Multiplying both the numerator and denominator of a(1 − rⁿ)/(1 − r) by −1 gives a(rⁿ − 1)/(r − 1). Convention: use whichever form keeps the numerator positive. When |r| < 1, use the first form. When r > 1, use the second form.

Variable Key

When to Use It

Use the sum formula when the total of a known number of terms is needed without adding them one by one. It applies to any compound-growth or repeated-multiplication problem with a finite cutoff — population doubling over a fixed period, loan interest over a known number of payments, or the total reach of a chain message over a set number of rounds.

The r = 1 Special Case

When r = 1, every term in the sequence equals the first term: a, a, a, … Plugging r = 1 into Sₙ = a(1 − rⁿ)/(1 − r) produces 0/0, which is undefined. The actual sum is just a added to itself n times, so:

Sₙ = na, when r = 1

This is why the Quick Reference table treats r = 1 as a separate row.

How the Formula Is Derived

The derivation uses an algebraic trick: subtracting Sₙ from rSₙ cancels every term except the first and last. Worked through with real numbers using a = 2, r = 3, n = 4:

Step 1. Write out the sum.

S₄ = 2 + 6 + 18 + 54

Step 2. Multiply both sides by r = 3.

3S₄ = 6 + 18 + 54 + 162

Step 3. Subtract the first equation from the second.

3S₄ − S₄ = (6 + 18 + 54 + 162) − (2 + 6 + 18 + 54)

3S₄ − S₄ = 162 − 2

2S₄ = 160

S₄ = 80

Verify by direct addition: 2 + 6 + 18 + 54 = 80. ✓

The same logic in symbols:

Sₙ = a + ar + ar² + … + ar^(n−1)

rSₙ = ar + ar² + … + ar^(n−1) + arⁿ

rSₙ − Sₙ = arⁿ − a

Sₙ(r − 1) = a(rⁿ − 1)

Sₙ = a(rⁿ − 1)/(r − 1)

This is where the formula comes from. Every term in the middle cancels.

Worked Example

Find the sum of the first 6 terms of 5, 10, 20, 40, …

a = 5, r = 2, n = 6

Since r > 1, use the second form:

S₆ = 5(2⁶ − 1)/(2 − 1) = 5(64 − 1)/1 = 5 · 63 = 315

Answer: S₆ = 315

Sum of an Infinite Geometric Sequence

The infinite sum formula:

S∞ = a/(1 − r), when |r| < 1

Diverges (no finite sum), when |r| ≥ 1

Variable Key

Why |r| < 1 Is Required

When |r| < 1, each term is smaller than the one before it. The terms shrink toward zero, and the partial sums approach a finite limit. When |r| ≥ 1, the terms either stay the same size (|r| = 1) or grow (|r| > 1), so adding infinitely many of them produces a sum that grows without bound or oscillates without settling.

The convergence condition follows directly from the finite-sum formula. As n grows, rⁿ approaches 0 when |r| < 1. Substituting that into Sₙ = a(1 − rⁿ)/(1 − r) gives:

S∞ = a(1 − 0)/(1 − r) = a/(1 − r)

Worked Example

Find the sum of the infinite series 4 + 2 + 1 + 0.5 + 0.25 + …

a = 4, r = 0.5, |r| < 1, so the series converges.

S∞ = 4/(1 − 0.5) = 4/0.5 = 8

Answer: S∞ = 8

Recursive vs Explicit Formulas

A geometric sequence can be written in two formula forms — explicit and recursive.

Use the explicit form when one specific term far in the sequence is needed. Use the recursive form when working through the sequence one term at a time, or when programming a loop.

Translation between the two forms is a Common Core standard requirement (CCSS HSF.BF.A.2) and is also covered in NCERT Class 11 Chapter 9.

Common Confusions

• Geometric vs arithmetic sequence. Geometric uses multiplication (× r each step); arithmetic uses addition (+ d each step). Mixing the two formula sets is the most common error in problems involving sequences.

• The two finite-sum forms. They are the same formula. Multiplying the numerator and denominator of (1 − rⁿ)/(1 − r) by −1 gives (rⁿ − 1)/(r − 1). Use whichever form keeps the numerator positive.

• n vs n − 1 in the exponent. The nth term uses r^(n−1), not r^n. The exponent counts how many times r has been multiplied — which is one less than the position. The first term has been multiplied by r zero times (r⁰ = 1).

• Infinite sum convergence. The infinite-sum formula only works when |r| < 1. Plugging in r = 2 or r = −1 produces a number, but that number is meaningless — the actual sum does not exist.

• First-term notation. Some textbooks use a, others a₁, others a₀ (indexing from zero). The values are identical, but the exponent shifts: aₙ = a · r^(n−1) when indexing from a₁, or aₙ = a · rⁿ when indexing from a₀.

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