A geometric sequence is a row of numbers made by multiplying by the same number each time.
If that multiplier, the common ratio r, is greater than 1 the terms snowball, if it is between 0 and 1 they keep shrinking, and if it is negative the signs alternate.
The nth term is the first term multiplied by the ratio (n − 1) times, and when the ratio's absolute value is less than 1, adding the terms forever still homes in on a single value.
In the first picture the bars grow or shrink with the first term and the ratio; in the second, the partial sums gather at the limit or run off the picture.
Geometric sequences: multiplying by the same ratio
seegongsik.com
Each bar is a term, with the first n terms shaded darker (the dotted line joins the tops of the bars)
2
2
5
💡 Neighboring terms always have the same ratio
①Divide any term by the one before it and you always get the common ratio r
②If r > 1, each step multiplies the term by r
③If 0 < r < 1, each step shrinks the term toward 0
④If r < 0, the signs alternate (this picture's ratio is at least 0.3; negative ratios can be chosen in the second picture)
The general term and the sum
General term
an = a₁ · rn-1
nth term = first term × ratioⁿ⁻¹
Sum of a geometric sequence
Sn = a₁(rn - 1)r - 1 (r ≠ 1)
If r = 1 every term is the same, so Sₙ = na₁
📐 Where the sum formula comes from
①Sₙ = a₁ + a₁r + a₁r² + … + a₁rⁿ⁻¹
②Multiplying by r gives rSₙ = a₁r + a₁r² + … + a₁rⁿ
③Subtracting wipes out all the middle terms and leaves Sₙ − rSₙ = a₁ − a₁rⁿ
④If r ≠ 1, dividing by 1 − r gives Sₙ = a₁(1 − rⁿ)/(1 − r)
Infinite geometric series: adding forever
seegongsik.com
Partial sums of a geometric series with first term 1. One term per row; each bar runs from the partial sum Sₙ to Sₙ₊₁
0.5
Infinite Geometric Series
S = a₁1 - r (|r| < 1)
It converges only when |r| < 1 and diverges when |r| ≥ 1. The first term in the picture is 1
Geometric Mean
Geometric Mean
b² = ac (a, b, c geometric)
The square of the middle term equals the product of the two outer terms
📐 Using the geometric mean
①Three nonzero numbers a, b, c form a geometric sequence in that order ⟺ b² = ac
②Since b = ±√(ac), the sign of b has to come from the conditions of the problem
③Terms picked at equal gaps from a geometric sequence form another geometric sequence
Summary
Geometric Core
an = a₁rn-1, Sn = a₁(rn - 1)r - 1
The general term and the sum side by side
🎯 Exam Points
①General term aₙ = a₁rⁿ⁻¹
②Sum Sₙ = a₁(rⁿ − 1)/(r − 1) (r ≠ 1), and Sₙ = na₁ when r = 1
③An infinite geometric series has S = a₁/(1 − r) when |r| < 1
④Geometric mean b² = ac
⑤The sign of the ratio decides whether signs alternate; its absolute value decides whether terms grow or shrink
Worked Examples & Past Exam
Example 1
Find the 5th term of a geometric sequence with first term 3 and common ratio 2.
1
Put n = 5 into the general term aₙ = a₁rⁿ⁻¹.
a5 = 3 × 25-1 = 3 × 24
2
2⁴ = 16, so multiply 3 by 16.
= 3 × 16 = 48
▸ a₅ = 48
Getting to the 5th term multiplies by the ratio four times. Listing 3, 6, 12, 24, 48 one by one gives the same value.
Example 2
Find the sum of the first 4 terms of a geometric sequence with first term 2 and common ratio 3.
1
Put a₁ = 2, r = 3, n = 4 into the sum formula Sₙ = a₁(rⁿ − 1)/(r − 1).
S4 = 2(34 - 1)3 - 1
2
3⁴ = 81, so the numerator is 2 × 80, and it is divided by the denominator 2.
= 2(81 - 1)2 = 80
▸ S₄ = 80
Adding 2, 6, 18, 54 directly also gives 80. The more terms there are, the more time the formula saves over adding one by one.
2022 KICE mock exam Math type, adapted
Find the sum of the infinite geometric series with first term 4 and common ratio 1/2.
①8
②4
③2
④6
⑤does not converge
▸ ① 8
1
|r| = 1/2 < 1, so the sum converges and S = a₁/(1 − r) applies.