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Determine the common ratio, the fifth term, and the \(n\) th term of the geometric sequence. $$t, \frac{t^{2}}{2}, \frac{t^{3}}{4}, \frac{t^{4}}{8}, \dots$$

Short Answer

Expert verified
Common ratio is \(\frac{t}{2}\), fifth term is \(\frac{t^5}{16}\), \(n\)th term is \(\frac{t^n}{2^{n-1}}\).

Step by step solution

01

Identify the First Term

The first term of the geometric sequence is given as the initial term, which is \( t \). This is denoted as \( a_1 = t \).
02

Calculate the Common Ratio

In a geometric sequence, the common ratio \( r \) is found by dividing any term by its preceding term. Calculate \( r \) using the first two terms: \[ r = \frac{\frac{t^2}{2}}{t} = \frac{t}{2} \].
03

Find the Fifth Term

The general formula for the \( n \)th term in a geometric sequence is \( a_n = a_1 \cdot r^{n-1} \). To find the fifth term, substitute \( a_1 = t \), \( r = \frac{t}{2} \), and \( n = 5 \): \[ a_5 = t \cdot \left(\frac{t}{2}\right)^{4} = t \cdot \frac{t^4}{16} = \frac{t^5}{16} \].
04

Derive the General Formula for the n-th Term

Using the general formula, substitute \( a_1 = t \) and the common ratio \( r = \frac{t}{2} \). This gives us the \( n \)th term: \[ a_n = t \left( \frac{t}{2} \right)^{n-1} = \frac{t^n}{2^{n-1}} \].

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Key Concepts

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

Understanding the Common Ratio
In a geometric sequence, the common ratio is a crucial component that defines how the sequence progresses from one term to the next. It is the factor by which each term is multiplied to get the subsequent term. To find this common ratio, simply take any term in the sequence, and divide it by its preceding term. This will remain consistent throughout the sequence. For the given sequence \( t, \frac{t^{2}}{2}, \frac{t^{3}}{4}, \frac{t^{4}}{8}, \ldots \), the common ratio is found by dividing the second term \( \frac{t^{2}}{2} \) by the first term \( t \). Thus,
  • \( r = \frac{\frac{t^2}{2}}{t} = \frac{t}{2} \).
The common ratio being \( \frac{t}{2} \) indicates that each subsequent term is the result of multiplying the previous term by \( \frac{t}{2} \). Understanding this ratio is the key to navigating through the terms of the sequence effectively.
Using the n-th Term Formula
The n-th term formula in a geometric sequence provides a direct way to compute any term in the sequence without calculating all the preceding terms. The formula is given by \( a_n = a_1 \cdot r^{n-1} \), where \( a_1 \) is the first term and \( r \) is the common ratio. For our sequence, with \( a_1 = t \) and \( r = \frac{t}{2} \), the n-th term, \( a_n \), can be calculated as:
  • \( a_n = t \cdot \left( \frac{t}{2} \right)^{n-1} = \frac{t^n}{2^{n-1}} \).
This simple formula allows us to find the position of any term in the sequence quickly. You don't have to manually multiply each term up to \( n \). Instead, plug \( n \) directly into the formula for an efficient computation.
Calculating the Fifth Term of the Sequence
Finding a specific term in a sequence, such as the fifth term, can be made straightforward using the n-th term formula. With the formula \( a_5 = a_1 \cdot r^{4} \) and knowing that \( a_1 = t \) and \( r = \frac{t}{2} \), we can compute the fifth term as follows:
  • \( a_5 = t \cdot \left( \frac{t}{2} \right)^4 = t \cdot \frac{t^4}{16} = \frac{t^5}{16} \).
This computation tells us that the fifth term of the sequence is \( \frac{t^5}{16} \). This method showcases how the n-th term formula simplifies the process, saving you time and ensuring precision.

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Most popular questions from this chapter

If the numbers \(a_{1}, a_{2}, \ldots, a_{n}\) form a geometric sequence, then \(a_{2}, a_{3}, \dots, a_{n-1}\) are geometric means between \(a_{1}\) and \(a_{n} .\) Insert three geometric means between 5 and 80 .

A construction company purchases a bulldozer for \(\$ 160,000\). Each year the value of the bulldozer depreciates by \(20 \%\) of its value in the preceding year. Let \(V_{n}\) be the value of the bulldozer in the \(n\) th year. (Let \(n=1\) be the year the bulldozer is purchased.) (a) Find a formula for \(V_{n}\). (b) In what year will the value of the bulldozer be less than \(\$ 100,000 ?\)

A ball is dropped from a height of \(9 \mathrm{ft}\). The elasticity of the ball is such that it always bounces up onethird the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fifth time. (b) Find a formula for the total distance the ball has traveled at the instant it hits the ground the \(n\) th time.

Find the term that does not contain \(x\) in the expansion of $$\left(8 x+\frac{1}{2 x}\right)^{8}$$.

A Recursively Defined Sequence Find the first 40 terms of the sequence defined by $$ a_{n+1}=\left\\{\begin{array}{ll} \frac{a_{n}}{2} & \text { if } a_{n} \text { is an even number } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is an odd number } \end{array}\right. $$ and \(a_{1}=11 .\) Do the same if \(a_{1}=25 .\) Make a conjecture about this type of sequence. Try several other values for \(a_{1},\) to test your conjecture.

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