Recursive Formula Geometric Sequence Explained

A common mistake is thinking a geometric sequence's common ratio can be any number. In reality, the ratio must be non-zero. A sequence with r = 0 is technically geometric but collapses to zero after the first term; a sequence with r = 1 is a constant sequence, still geometric. The recursive formula geometric sequence uses the previous term to find the next, while the explicit formula gives any term directly from its position.

The recursive form defines a term using the one before it. For a geometric sequence, the rule is aₙ = r * aₙ₋₁, where r is the common ratio and a₁ is given. This form is straightforward for generating terms sequentially but requires knowing the previous term.

For example, if a₁ = 3 and r = 2, the recursive rule aₙ = 2 * aₙ₋₁ produces the sequence: 3, 6, 12, 24, 48, ... Each term after the first is found by multiplying the previous term by 2. The recursive form is ideal for problems that ask for the next few terms or for understanding the multiplicative relationship.

OpenStax Algebra and Trigonometry 2e (section 13.3, pages 827-838) defines the recursive formula as aₙ = r * aₙ₋₁, with r ≠ 0. The domain of n is positive integers, meaning a₁ is the starting point.

The explicit form gives the n-th term directly using the index: aₙ = a₁ * r^(n-1). This formula lets you find any term without computing all previous ones. Use it for solving problems where you need the term far into the sequence or the term count from a given value.

Using the same example, a₁ = 3 and r = 2, the explicit formula is aₙ = 3 * 2^(n-1). For n = 5, a₅ = 3 * 2^(4) = 3 * 16 = 48, matching the recursive output. The explicit form comes from OpenStax Algebra and Trigonometry 2e (section 13.3), and is the standard for a geometric sequence's n-th term.

To convert a recursive form to explicit, identify a₁ and r from the recursive rule aₙ = r * aₙ₋₁. Then write aₙ = a₁ * r^(n-1). For example, given a₁ = 5 and aₙ = 3 * aₙ₋₁, the explicit form is aₙ = 5 * 3^(n-1).

To go from explicit to recursive, extract a₁ from the explicit formula (the coefficient) and r from the base of the exponent. For aₙ = 7 * 4^(n-1), a₁ = 7 and r = 4, so the recursive form is a₁ = 7, aₙ = 4 * aₙ₋₁ for n ≥ 2. This conversion is a common Algebra 1/2 task.

Common Failure: Sign Error in r

When converting from two non-consecutive terms, finding r requires r = (aₙ / aₘ)^(1/(n-m)). If n-m is even and the ratio of terms is negative, taking the root can lose the sign. For example, given a₂ = 8 and a₄ = 32, r = (32/8)^(1/(4-2)) = (4)^(1/2) = ±2. Both 2 and -2 produce valid sequences, but the problem context (e.g., positive terms) determines which to use. Always check if terms alternate sign, which indicates a negative r.

Failure: Logarithm of a Negative Number

When solving for n from a partial sum Sₙ = a₁(1 - rⁿ)/(1 - r), the argument of the logarithm can become negative if a₁ and (Sₙ * (1 - r)) have opposite signs. For instance, with a₁ = 10, r = 2, and Sₙ = 30, the equation yields a negative argument, meaning no real solution for n exists. Use the separate formula Sₙ = n * a₁ only when r = 1, and verify the sign before solving.

The recursive form is useful for generating terms step by step, especially in computer programs or when modeling processes where each step depends on the previous one. It is also easier for understanding the multiplicative nature of geometric sequences.

The explicit form is better for finding a specific term quickly, for solving for n or r from given terms, and for calculating series sums. For infinite geometric series, the explicit form directly leads to the sum formula S∞ = a₁/(1 - r), valid only when |r| < 1, as stated in Stewart Calculus (section 11.2, pages 716-726).

Use the recursive form when you have a₁ and r and need the next few terms, or when the problem gives a term index and asks for the next. Use the explicit form when you need a term far into the sequence or when working with the sum formulas, including the partial sum Sₙ = a₁(1 - rⁿ)/(1 - r) for r ≠ 1 (OpenStax, section 13.4, pages 839-850).

For r = 1, the partial sum formula is Sₙ = n * a₁, a separate edge case. Do not use the general formula when r = 1, as it is undefined (division by zero).

Work through these problems to convert between recursive and explicit forms. Answers are provided.

Problem 1

A geometric sequence has a₁ = 4 and common ratio r = -3. Write the recursive form and the explicit form. Then find a₄.

Answer: Recursive: a₁ = 4, aₙ = -3 * aₙ₋₁. Explicit: aₙ = 4 * (-3)^(n-1). a₄ = 4 * (-3)^(3) = 4 * (-27) = -108.

Problem 2

Given the explicit formula aₙ = 5 * 2^(n-1), write the recursive form. Then find the 6th term.

Answer: Recursive: a₁ = 5, aₙ = 2 * aₙ₋₁. a₆ = 5 * 2^(5) = 5 * 32 = 160.

Problem 3

Convert the recursive rule a₁ = 12, aₙ = 0.5 * aₙ₋₁ to explicit form. Is the sequence converging? If so, find S∞.

Answer: Explicit: aₙ = 12 * (0.5)^(n-1). Since |r| = 0.5 < 1, the sequence converges. S∞ = a₁/(1 - r) = 12/(1 - 0.5) = 12/0.5 = 24.

Problem 4

A geometric sequence has a₃ = 18 and a₆ = 486. Find r and a₁, then write both forms.

Answer: r = (a₆/a₃)^(1/(6-3)) = (486/18)^(1/3) = 27^(1/3) = 3. Then a₁ = a₃ / r^(3-1) = 18 / 3^2 = 18/9 = 2. Explicit: aₙ = 2 * 3^(n-1). Recursive: a₁ = 2, aₙ = 3 * aₙ₋₁.

Problem 5

For the geometric sequence with r = -1 and a₁ = 7, write both forms and find S₅ using the appropriate formula. Does the infinite sum exist?

Answer: Explicit: aₙ = 7 * (-1)^(n-1). Recursive: a₁ = 7, aₙ = -1 * aₙ₋₁. For r = -1, use Sₙ = a₁(1 - rⁿ)/(1 - r) since r ≠ 1. S₅ = 7(1 - (-1)^5)/(1 - (-1)) = 7(1 - (-1))/(2) = 7(2)/2 = 7. The infinite sum does not exist because |r| = 1, not < 1; the series diverges (Stewart Calculus, section 11.2).