Question 1

How do I find the common ratio of a geometric sequence?

Accepted Answer

Divide any term by the previous term: r = aₙ / aₙ₋₁. Example: in 5, 15, 45, 135, … r = 15/5 = 3. If you know two non-consecutive terms: r = (aₘ / aₙ)^(1/(m−n)). Example: a₂ = 8 and a₅ = 64 → r = (64/8)^(1/3) = 8^(1/3) = 2.

Question 2

When does an infinite geometric series have a finite sum?

Accepted Answer

An infinite geometric series converges (has a finite sum) only when |r| < 1. As n → ∞, rⁿ → 0, so S∞ = a₁ / (1 − r). Example: 1 + 1/2 + 1/4 + 1/8 + … = 1 / (1 − 0.5) = 2. If |r| ≥ 1, the terms do not shrink, so the series diverges — the sum grows without bound.

Question 3

What is the difference between geometric and arithmetic sequences?

Accepted Answer

Arithmetic sequences add a constant d (linear growth): aₙ = a₁ + (n−1)d. Geometric sequences multiply by a constant r (exponential growth): aₙ = a₁ · rⁿ⁻¹. Plotting an arithmetic sequence gives a straight line; a geometric sequence gives an exponential curve. Investment compounding is geometric (interest on interest); equal monthly savings deposits create an arithmetic series.

Question 4

How is compound interest related to geometric sequences?

Accepted Answer

Compound interest is a geometric sequence where each term is the balance after one compounding period. If you invest P at rate r per period, the balance after n periods is Pₙ = P · (1+r)ⁿ — exactly the nth term of a geometric sequence with a₁ = P(1+r) and ratio (1+r). The total interest earned over n periods is the geometric series sum minus the principal.

Question 5

What happens when r is negative in a geometric sequence?

Accepted Answer

When r < 0, the sequence alternates in sign: positive, negative, positive, negative … Example: 4, −8, 16, −32 … (r = −2). The terms alternate around zero but grow in absolute value if |r| > 1, or shrink if |r| < 1. An alternating geometric series with |r| < 1 still converges to a₁/(1−r), which will be a value between 0 and a₁.

Question 6

How do I find the number of terms given the first term, ratio, and last term?

Accepted Answer

Use aₙ = a₁ · rⁿ⁻¹ and solve for n: rⁿ⁻¹ = aₙ/a₁ → (n−1) = log(aₙ/a₁) / log(r) → n = 1 + log(aₙ/a₁) / log(r). Example: first term 3, ratio 2, last term 192: n = 1 + log(64) / log(2) = 1 + 6 = 7. The sequence is 3, 6, 12, 24, 48, 96, 192 — indeed 7 terms.

Formula	Expression	Description
nth term	aₙ = a₁ · rⁿ⁻¹	Any term given first term and ratio
Common ratio	r = aₙ / aₙ₋₁	Constant multiplier between consecutive terms
Partial sum (r≠1)	Sₙ = a₁(1 − rⁿ) / (1 − r)	Sum of first n terms
Partial sum (r=1)	Sₙ = n × a₁	All terms equal — just multiply
Infinite sum	S∞ = a₁ / (1 − r)	Only valid when \|r\| < 1
Find n	n = log(aₙ/a₁) / log(r) + 1	Which position has value aₙ?
Geometric mean	GM = √(a × b)	Middle term between a and b

Sequence	r	a₈	S₈	S∞
Powers of 2: 1,2,4,8,…	2	128	255	Diverges
Half each time: 1, ½, ¼,…	0.5	1/128 ≈ 0.0078	≈ 1.992	2
Powers of 3: 1,3,9,27,…	3	2187	3280	Diverges
2/3 decay: 1, 2/3, 4/9,…	0.667	≈ 0.039	≈ 2.961	3
−½ alternating: 1,−½,¼,…	−0.5	≈ 0.0039	≈ 0.668	2/3 ≈ 0.667
Compound 5%: 1000·1.05ⁿ	1.05	1477.46	9549.11	Diverges

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