Math Problem Statement

I. Find the sum of the following geometric sequence:

  1. 1, 2, 4, 8, ..., A12
  2. 3, 6, 12, 24, ..., A8
  3. 512, 256, 128, ..., A6
  4. 12, 18, 27, ..., A9
  5. 2, 10, 50, ..., A6

Solution

To find the sum of each geometric sequence, we use the sum formula for a geometric sequence:

Sn=a1rn1rifr<1S_n = a \cdot \frac{1 - r^n}{1 - r} \quad \text{if} \quad |r| < 1 or Sn=arn1r1ifr>1S_n = a \cdot \frac{r^n - 1}{r - 1} \quad \text{if} \quad r > 1 where:

  • SnS_n is the sum of the first nn terms,
  • aa is the first term of the sequence,
  • rr is the common ratio, and
  • nn is the number of terms.

1. Sequence: 1, 2, 4, 8, ..., A12

  • First term: a=1a = 1
  • Common ratio: r=2r = 2
  • Number of terms: n=12n = 12

The sum is:

S12=1212121=2121=40961=4095S_{12} = 1 \cdot \frac{2^{12} - 1}{2 - 1} = 2^{12} - 1 = 4096 - 1 = 4095

2. Sequence: 3, 6, 12, 24, ..., A8

  • First term: a=3a = 3
  • Common ratio: r=2r = 2
  • Number of terms: n=8n = 8

The sum is:

S8=328121=3(2561)=3255=765S_8 = 3 \cdot \frac{2^8 - 1}{2 - 1} = 3 \cdot (256 - 1) = 3 \cdot 255 = 765

3. Sequence: 512, 256, 128, ..., A6

  • First term: a=512a = 512
  • Common ratio: r=12r = \frac{1}{2}
  • Number of terms: n=6n = 6

The sum is:

S6=5121(12)6112=512116412=512636412=5126332=1008S_6 = 512 \cdot \frac{1 - \left(\frac{1}{2}\right)^6}{1 - \frac{1}{2}} = 512 \cdot \frac{1 - \frac{1}{64}}{\frac{1}{2}} = 512 \cdot \frac{\frac{63}{64}}{\frac{1}{2}} = 512 \cdot \frac{63}{32} = 1008

4. Sequence: 12, 18, 27, ..., A9

  • First term: a=12a = 12
  • Common ratio: r=32r = \frac{3}{2}
  • Number of terms: n=9n = 9

The sum is:

S9=12(32)91321=12(1.5)910.5S_9 = 12 \cdot \frac{\left(\frac{3}{2}\right)^9 - 1}{\frac{3}{2} - 1} = 12 \cdot \frac{(1.5)^9 - 1}{0.5} S9=1238.44335937510.5=1237.4433593750.5=1274.88671875=898.640625S_9 = 12 \cdot \frac{38.443359375 - 1}{0.5} = 12 \cdot \frac{37.443359375}{0.5} = 12 \cdot 74.88671875 = 898.640625

5. Sequence: 2, 10, 50, ..., A6

  • First term: a=2a = 2
  • Common ratio: r=5r = 5
  • Number of terms: n=6n = 6

The sum is:

S6=256151=21562514=2156244=23906=7812S_6 = 2 \cdot \frac{5^6 - 1}{5 - 1} = 2 \cdot \frac{15625 - 1}{4} = 2 \cdot \frac{15624}{4} = 2 \cdot 3906 = 7812

Summary:

  1. S12=4095S_{12} = 4095
  2. S8=765S_8 = 765
  3. S6=1008S_6 = 1008
  4. S9898.64S_9 \approx 898.64
  5. S6=7812S_6 = 7812

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

  1. How does changing the common ratio affect the sum of a geometric series?
  2. Can the sum of a geometric series be infinite?
  3. How does the sum change if the first term is negative?
  4. What is the difference between a geometric sequence and an arithmetic sequence?
  5. How do we find the sum of a geometric series with a common ratio less than 1?

Tip: When the common ratio is a fraction between 0 and 1, the terms of the geometric sequence decrease rapidly, leading to a smaller sum.

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Math Problem Analysis

Mathematical Concepts

Geometric Sequences
Series Summation

Formulas

Sum of geometric series: S_n = a * (r^n - 1) / (r - 1) for r > 1
Sum of geometric series: S_n = a * (1 - r^n) / (1 - r) for |r| < 1

Theorems

Geometric Series Sum Theorem

Suitable Grade Level

Grades 9-12