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Chapter 13: Problem 24

Find the indicated term for each geometric sequence $$ a_{1}=1, r=3 ; \quad a_{15} $$

Short Answer

Expert verified
The 15th term \( a_{15} \) of the geometric sequence is 4782969.

Step by step solution

01

Identify the general formula for a geometric sequence

The general formula for the nth term of a geometric sequence is given by a_{n} = a_{1} \times r^{n-1}where \( a_{n} \) is the nth term, \( a_{1} \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
02

Plug in the known values

We are given \( a_{1}=1 \), \( r=3 \), and need to find \( a_{15} \). Substitute these values into the general formula:a_{15} = 1 \times 3^{15-1}
03

Simplify the exponent

Calculate the exponent part of the formula:3^{15-1} = 3^{14}
04

Compute the power

Calculate \( 3^{14} \):3^{14} = 4782969
05

Multiply with the first term

Lastly, multiply the result by the first term \( a_{1} \):a_{15} = 1 \times 4782969 = 4782969

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

nth term
The nth term in a geometric sequence is crucial for identifying any term within the sequence. To find it, we rely on a standard formula: \[ a_{n} = a_{1} \times r^{n-1} \]This formula allows us to determine any term if we know the first term (\( a_{1} \)), the common ratio (\( r \)), and the term number (\( n \)). Each variable in the formula has a specific role:
  • The first term (\( a_{1} \) ) is the initial term of the sequence.
  • The common ratio (\( r \)) shows how each term relates to the previous one.
  • The term number (\( n \)) tells us which term we are trying to find.
Plugging these values into the formula gives us the desired term in the geometric sequence.
common ratio
The common ratio (\( r \)) is a key concept in geometric sequences. It defines the constant factor by which we multiply each term to get the next term. Let's illustrate this:
  • If the sequence starts with 2 and has a common ratio of 3, the next terms will be: 2, 6, 18, 54, etc. This sequence multiplies by 3 each time.
  • Similarly, if the common ratio is 1/2 and the first term is 8, the sequence would look like: 8, 4, 2, 1, etc. Each term is half of the previous one.
Clearly understanding the common ratio helps us see how the sequence progresses and gives insight into its growth pattern. The common ratio can be found by dividing any term by the previous term in the sequence.
exponentiation
Exponentiation is the process of raising a number to a power, which is fundamental in calculating terms of a geometric sequence. In the context of our formula, \(a_{n} = a_{1} \times r^{n-1}\), exponentiation is used to elevate the common ratio (\( r \)) to the power of (\( n-1 \)).
This power operation allows us to scale the common ratio appropriately for reaching the nth term. For example, in our given problem, we calculated \(3^{14}\):
  • The base (3) is the common ratio, indicating each term is three times the previous term.
  • The exponent (14) comes from \( n-1 \) and scales up 3 step-by-step to the 15th term.
Understanding exponentiation is essential as it directly impacts the magnitude of the terms in the sequence, especially when working with high values like \(3^{14}\) which equals 4,782,969.
sequence formula
The sequence formula for a geometric sequence is the backbone of solving these problems. It is given by: \[ a_{n} = a_{1} \times r^{n-1} \] This equation puts together the first term (\( a_{1} \)), the common ratio (\( r \)), and the term position (\( n \)).
Using this formula helps us find any term without having to manually multiply the common ratio repeatedly. For instance, in our exercise, we needed the 15th term:
  • We knew the first term \( a_{1} \) was 1.
  • The common ratio (\( r \)) was 3.
  • We wanted the 15th term, so \( n=15 \)
Plugging these into the sequence formula, we computed \(1 \times 3^{14} = 4782969\). This formula is powerful for efficiently finding terms in any geometric sequence.

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