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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblem 10.R.95a
Chapter 10, Problem 10.R.95a

Building a tunnel — first scenario
A crew of workers is constructing a tunnel through a mountain. Understandably, the rate of construction decreases because rocks and earth must be removed a greater distance as the tunnel gets longer. Suppose each week the crew digs 0.95 of the distance it dug the previous week. In the first week, the crew constructed 100 m of tunnel.


a.How far does the crew dig in 10 weeks? 20 weeks? N weeks?

Verified step by step guidance
1
Identify the type of sequence described in the problem. Since each week the crew digs 0.95 times the distance dug the previous week, this is a geometric sequence where each term is multiplied by a common ratio \(r = 0.95\).
Write the general formula for the distance dug in the \(n^{th}\) week. The first term \(a_1\) is 100 meters, so the distance dug in week \(n\) is given by \(a_n = a_1 \times r^{n-1} = 100 \times 0.95^{n-1}\).
To find the total distance dug after \(N\) weeks, recognize that this is the sum of the first \(N\) terms of a geometric sequence. Use the formula for the sum of the first \(N\) terms: \(S_N = a_1 \times \frac{1 - r^N}{1 - r}\).
Apply the sum formula to calculate the total distance dug after 10 weeks by substituting \(N=10\), \(a_1=100\), and \(r=0.95\) into the formula: \(S_{10} = 100 \times \frac{1 - 0.95^{10}}{1 - 0.95}\).
Similarly, calculate the total distance dug after 20 weeks by substituting \(N=20\) into the sum formula: \(S_{20} = 100 \times \frac{1 - 0.95^{20}}{1 - 0.95}\). For a general number of weeks \(N\), use \(S_N = 100 \times \frac{1 - 0.95^{N}}{1 - 0.95}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Geometric Sequences

A geometric sequence is a list of numbers where each term is found by multiplying the previous term by a constant ratio. In this problem, the distance dug each week decreases by a factor of 0.95, making it a geometric sequence with first term 100 and common ratio 0.95.
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Geometric Sequences - Recursive Formula

Sum of a Finite Geometric Series

The total distance dug over multiple weeks is the sum of the terms in the geometric sequence. The sum of the first n terms of a geometric series can be calculated using the formula S_n = a(1 - r^n) / (1 - r), where a is the first term and r is the common ratio.
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Generalization to N Terms

To find the distance dug after N weeks, the formula for the sum of the first N terms of the geometric series is used. This allows calculation of the total distance for any number of weeks, not just specific values like 10 or 20.
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Related Practice
Textbook Question

Explain why or why not

Determine whether the following statements are true and give an explanation or counterexample.


g.The series ∑ (from k = 1 to ∞) (k² / (k² + 1)) converges.

Textbook Question

89–90. {Use of Tech} Lower and upper bounds of a series

For each convergent series and given value of n, complete the following.


b. Find an upper bound for the remainder Rₙ.


89.∑ (from k = 1 to ∞)1 / k⁵ ;n = 5

Textbook Question

Determine whether the following statements are true and give an explanation or counterexample.

c. The terms of the sequence of partial sums of the series ∑ aₖ approach 5/2, so the infinite series converges to 5/2.

Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)k√k / k³

Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)k! / (eᵏ kᵏ)

Textbook Question

89–90. {Use of Tech} Lower and upper bounds of a series

For each convergent series and given value of n, complete the following.


c. Find lower and upper bounds (Lₙ and Uₙ respectively) for the exact value of the series.


89.∑ (from k = 1 to ∞)1 / k⁵ ;n = 5