☰ Contents · Algebra

Geometric progression

Lessons 32 · 1 lessons · Sh.A. Alimov, A.R. Xalmukhamedov, M.A. Mirzaakhmedov. Algebra: textbook for Grade 9 of general secondary schools. Revised 4th edition. Tashkent: “O‘qituvchi”, 2019
32

Geometric progression

Textbook: pp. 162–166
GoalKnow the definition of a geometric progression, its common ratio and the n-th term formula; apply them in problems.
New words
geometric progression · geometrik progressiyacommon ratio · progressiya maxrajigeometric mean · o‘rta geometrikn-th term formula · n-had formulasi
Explanation

A sequence with first term b₁ ≠ 0 in which each term from the second is the previous term multiplied by the same number q ≠ 0 is called a geometric progression: b_(n+1) = b_n · q. The number q is the common ratio: q = b_(n+1) : b_n. For example, in 3, 6, 12, 24, ... q = 2; in 8, 4, 2, 1, ... q = 1/2; in 5, −10, 20, ... q = −2. To prove that a sequence is a geometric progression we show that b_(n+1) : b_n does not depend on n. The n-th term formula is b_n = b₁ · q^(n−1). In a progression with all terms positive, b_n² = b_(n−1) · b_(n+1), i.e. each term is the geometric mean of its neighbours: b_n = √(b_(n−1) b_(n+1)); this gives the name. If two terms are given, dividing them gives a power of q.

Worked examples
3, 6, 12, ...: q = 2, b₆ = 3 · 2⁵ = 96. Is 1458 a term of 2, 6, 18, ...? 2 · 3^(n−1) = 1458, 3^(n−1) = 729 = 3⁶, so n = 7, yes.
b₃ = 12, b₆ = 96: b₆ : b₃ = q³ = 8, so q = 2; b₁ = b₃ : q² = 12 : 4 = 3. Formula: b_n = 3 · 2^(n−1).
Class activity

“Multiplier chain”: one student states b₁ and q, the class takes turns computing b₂, b₃, b₄; then the reverse: terms are given and q must be found.

Practice
1
Find the common ratio of 5, −10, 20, ...
2
If b₁ = 4 and q = 3, find b₅.
3
Is the sequence b_n = 5 · 2ⁿ a geometric progression?
4
Why is b_n = √(b_(n−1) b_(n+1)) in a progression with positive terms?