☰ Contents · Algebra

Infinite decreasing progression

Lessons 34 · 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
34

Infinite decreasing geometric progression

Textbook: pp. 171–176
GoalRecognise an infinite decreasing geometric progression, find its sum and turn a periodic decimal into an ordinary fraction.
New words
infinite decreasing progression · cheksiz kamayuvchi progressiyalimit · limitinfinite sum · cheksiz yig‘indiperiodic decimal · davriy kasr
Explanation

A geometric progression whose common ratio has modulus less than one, i.e. |q| < 1, is called infinite decreasing; its terms get closer to zero as n grows. If n grows without bound, q^n tends to zero, which we write lim q^n = 0 (n → ∞). So in S_n = b₁ : (1 − q) − b₁q^n : (1 − q) the second part tends to zero, while the first does not depend on n. The sum of an infinite decreasing progression is the limit of S_n as n → ∞, and it equals S = b₁ : (1 − q). This formula is valid only for |q| < 1. We also use it to turn a periodic decimal into an ordinary fraction: 0.(7) = 7/10 + 7/100 + ... .

Worked examples
8 + 4 + 2 + ...: b₁ = 8, q = 1/2, |q| < 1. S = 8 : (1 − 1/2) = 8 : 1/2 = 16.
0.(7) = 7/10 + 7/100 + 7/1000 + ...: b₁ = 7/10, q = 1/10. S = (7/10) : (9/10) = 7/9.
Class activity

“Halving paper”: shade half of a square sheet, then half of what remains; write the shaded areas as 1/2, 1/4, 1/8 ... and discuss what their sum approaches.

Practice
1
Is 3, 1, 1/3, ... an infinite decreasing progression?
2
Find 18 + 6 + 2 + ...
3
Write the periodic decimal 0.(6) as an ordinary fraction.
4
Why can S = b₁ : (1 − q) not be used for 1 + 2 + 4 + ... ?