IB Mathematics AA SL The sum of an infinite geometric sequence Study Notes
IB Mathematics AA SL The sum of an infinite geometric sequence Study Notes Offer a clear explanation of The sum of an infinite geometric sequence , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic The sum of an infinite geometric sequence
Sum of an Infinite Geometric Sequence
Sum of Infinite Convergent Geometric Sequences
A geometric sequence is a sequence where each term is found by multiplying the previous term by a constant ratio \( r \).
When a geometric sequence continues infinitely and the common ratio satisfies \( |r| < 1 \), the sequence is convergent and the sum approaches a finite value.
Condition for Convergence
A geometric sequence with first term \( a \) and common ratio \( r \) converges if and only if:
\( |r| < 1 \)
If \( |r| \geq 1 \), the series does not converge (i.e., the sum is infinite or undefined).
Formula for the Sum of an Infinite Geometric Series:
If \( |r| < 1 \), then the sum to infinity is:
\( S = \frac{a}{1 – r} \)
- \( S \): Sum to infinity
- \( a \): First term of the series
- \( r \): Common ratio
Examples
Find the sum to infinity for the following geometric series (if it converges):
- \( 5 + 2.5 + 1.25 + \ldots \)
- \( 12 – 8 + \frac{16}{3} – \ldots \)
- \( \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots \)
▶️ Answer/Explanation
1. First term \( a = 5 \), common ratio \( r = \frac{2.5}{5} = 0.5 \)
Since \( |r| = 0.5 < 1 \), it converges.
\( S = \frac{5}{1 – 0.5} = \frac{5}{0.5} = 10 \)
2. First term \( a = 12 \), common ratio \( r = \frac{-8}{12} = -\frac{2}{3} \)
Since \( |r| = \frac{2}{3} < 1 \), it converges.
\( S = \frac{12}{1 – (-\frac{2}{3})} = \frac{12}{1 + \frac{2}{3}} = \frac{12}{\frac{5}{3}} = \frac{36}{5} = 7.2 \)
3. First term \( a = \frac{9}{10} \), ratio \( r = \frac{1}{10} \)
Since \( |r| = \frac{1}{10} < 1 \), it converges.
\( S = \frac{\frac{9}{10}}{1 – \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1 \)