Math Problem Statement

Good morning sa lahat. Kindly answer the question below.(Notebook).Sundan nio lang yung naisend kong topic kahapon. Find the sum to infinity of each geometric sequence.

  1. 64,16, 4,1...

  2. 1/3, 1/9, 1/27, 1/81...

Solution

Good morning! Let's solve each of the geometric sequence problems one by one.

Formula for the Sum to Infinity of a Geometric Sequence:

S=a1rS_{\infty} = \frac{a}{1 - r} where:

  • SS_{\infty} is the sum to infinity,
  • aa is the first term of the sequence,
  • rr is the common ratio, and r<1|r| < 1.

1) Sequence: 64,16,4,1,64, 16, 4, 1, \dots

  • First term (aa) = 64
  • Common ratio (rr) = 1664=14\frac{16}{64} = \frac{1}{4}

Now, applying the sum to infinity formula:

S=64114=6434=64×43=2563S_{\infty} = \frac{64}{1 - \frac{1}{4}} = \frac{64}{\frac{3}{4}} = 64 \times \frac{4}{3} = \frac{256}{3}

Thus, the sum to infinity is 2563\frac{256}{3} or approximately 85.33.

2) Sequence: 13,19,127,\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots

  • First term (aa) = 13\frac{1}{3}
  • Common ratio (rr) = 1913=13\frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}

Applying the sum to infinity formula:

S=13113=1323=12S_{\infty} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}

Thus, the sum to infinity is 12\frac{1}{2}.


Would you like further details or have any questions?

Here are some related questions to explore:

  1. What if the common ratio rr is greater than 1?
  2. How do you find the sum of a finite geometric sequence?
  3. Can the sum to infinity exist if r1|r| \geq 1?
  4. What is the sum to infinity when the first term is negative?
  5. How do geometric series relate to exponential decay?

Tip: Always check if the absolute value of the common ratio r|r| is less than 1, otherwise the sum to infinity does not exist!

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Math Problem Analysis

Mathematical Concepts

Geometric Sequences
Sum to Infinity
Common Ratio

Formulas

Sum to Infinity Formula: S_{∞} = a / (1 - r)

Theorems

Geometric Series Convergence Theorem

Suitable Grade Level

Grades 9-11