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?
Practice Questions
1 question
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
As n approaches infinity and |r| < 1, r^n approaches 0, thus S_n approaches a/(1-r).
Questions & Step-by-step Solutions
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Q
Q: 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?
Solution: As n approaches infinity and |r| < 1, r^n approaches 0, thus S_n approaches a/(1-r).
