Math Problem Statement

Find the 8th term of the following geometric sequence: 4, 12, 36, 108, ...

Solution

The problem asks to find the 8th term of a given geometric sequence:

Sequence: 4,12,36,108,4, 12, 36, 108, \dots

We are given:

  • First term (aa) = 4
  • Common ratio (rr) = 124=3\frac{12}{4} = 3
  • Term number (nn) = 8

In a geometric sequence, the general term (ana_n) is calculated using the formula: an=arn1a_n = a \cdot r^{n-1} Substituting the given values: a8=4381=437a_8 = 4 \cdot 3^{8-1} = 4 \cdot 3^7 Now, calculate 373^7: 37=21873^7 = 2187 So: a8=42187=8748a_8 = 4 \cdot 2187 = 8748 Thus, the 8th term of the sequence is 87488748.

Would you like further details or explanations?

Here are 5 related questions to deepen your understanding:

  1. How would you find the 5th term of this sequence?
  2. What is the sum of the first 8 terms of this geometric sequence?
  3. How can you determine if a sequence is arithmetic or geometric?
  4. What happens if the common ratio is less than 1? How would that affect the terms?
  5. How does the formula for the nth term of a geometric sequence compare to that of an arithmetic sequence?

Tip: Always verify the common ratio by dividing consecutive terms to ensure the sequence is geometric.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometric Sequences
Exponentiation
Patterns in Sequences

Formulas

a_n = a * r^(n-1)

Theorems

Formula for nth term of a geometric sequence

Suitable Grade Level

Grades 8-10