Missing terms of a geometric seqeunce

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SUMMARY

The discussion focuses on solving a geometric sequence problem where the first five terms are given as 3, __, 32x+1, __, __. The formula used is tn=arn-1, leading to the equation 32x+1=3r3-1. The solution reveals that the common ratio r can be simplified to r=√(32x+1/3), which further simplifies to 3x after applying the hint regarding the relationship between consecutive terms in a geometric sequence.

PREREQUISITES
  • Understanding of geometric sequences and their properties
  • Familiarity with the formula for the nth term of a geometric sequence (tn=arn-1)
  • Basic algebraic manipulation, including simplification of fractions
  • Knowledge of exponents and roots, particularly with base 3
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  • Study the properties of geometric sequences in detail
  • Practice solving problems involving the nth term of geometric sequences
  • Learn about the relationship between consecutive terms in geometric sequences
  • Explore advanced topics in sequences and series, including convergence and divergence
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Students studying algebra, particularly those focusing on sequences and series, as well as educators looking for examples to illustrate geometric sequences.

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Homework Statement


Write the first 5 terms of the geometric sequence
3, __ , 32x+1, __, __

Homework Equations


tn=arn-1


The Attempt at a Solution


tn=arn-1
32x+1=3r3-1
r2=32x+1 / 3
r=√32x+1 / √3

I'm stuck
 
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Before taking the square root, try to simplyfy the right side. What is \frac{3^{2x+1}}{3^{1}}?
 
Hint:

If a,b,c are three consecutive terms of a geometric sequence, b^2 = ac.
 
Villyer said:
Before taking the square root, try to simplyfy the right side. What is \frac{3^{2x+1}}{3^{1}}?

Ohh I see. 32x+1 / 31 = 32x. And the square root of that is 3x.
Thank you
 

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