If the sum of the first n terms of a geometric progression is given by S_n = a(1
Practice Questions
Q1
If the sum of the first n terms of a geometric progression is given by S_n = a(1 - r^n) / (1 - r), what happens to S_n as n approaches infinity when |r| < 1?
S_n approaches 0
S_n approaches infinity
S_n approaches a/(1-r)
S_n approaches a
Questions & Step-by-Step Solutions
If the sum of the first n terms of a geometric progression is given by S_n = a(1 - r^n) / (1 - r), what happens to S_n as n approaches infinity when |r| < 1?
Step 1: Understand the formula for the sum of the first n terms of a geometric progression: S_n = a(1 - r^n) / (1 - r).
Step 2: Identify the condition given in the question: |r| < 1. This means that the absolute value of r is less than 1.
Step 3: Recognize what happens to r^n as n becomes very large (approaches infinity). Since |r| < 1, r^n gets smaller and smaller, approaching 0.
Step 4: Substitute the limit of r^n into the formula for S_n. As n approaches infinity, r^n approaches 0, so S_n becomes S_n = a(1 - 0) / (1 - r).
Step 5: Simplify the expression: S_n = a / (1 - r). This is the value S_n approaches as n approaches infinity.
