Question

There is a strange journey appointed to a man. The first day he must go \(1 \frac{1}{2}\) miles, and every day after the first he must increase his journey by \(\frac{1}{6}\) of a mile, so that his journey shall proceed by an arithmetical progression. And he has to travel for his whole journey 2955 miles. In what number of days will he end his journey?

Short answer

Answer: 77 days

Step-by-step solution

Set up the arithmetic series formula

The distance traveled each day forms an arithmetic series. The general formula for the sum of an arithmetic series is given by: \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(S_n\) is the total distance, \(n\) is the number of terms (days), \(a_1\) is the distance traveled on the first day, and \(a_n\) is the distance traveled on the final day.

Write down the given values and find the last term formula

In our case, we have the following values: \(S_n = 2955\) miles \(a_1 = 1 \frac{1}{2} = \frac{3}{2}\) miles Since the common difference \(d = \frac{1}{6}\) mile, and \(a_n = a_1 +(n-1)d\), we can write the last term formula as: \(a_n = \frac{3}{2} +(n-1)\frac{1}{6}\)

Substitute the values in the arithmetic series formula and solve for n

Plug the values of \(S_n\), \(a_1\), and \(a_n\) into the arithmetic series formula: \(2955 = \frac{n}{2}\Big(\frac{3}{2} + \frac{3}{2} +(n-1)\frac{1}{6}\Big)\) Now we solve the equation for \(n\): \(2955 = \frac{n}{2}\Big(\frac{6}{2} + (n-1)\frac{1}{6}\Big)\) \(2955 = \frac{n}{2}\Big(\frac{6 + 6(n-1)}{6}\Big)\) \(2955 = \frac{n}{2}\Big(\frac{6 + 6n - 6}{6}\Big)\) \(2955 = \frac{n}{2}(n)\) \(2955 = \frac{n^2}{2}\) Multiply both sides by 2 to get rid of the fraction: \(5910 = n^2\) Take the square root of both sides: \(\sqrt{5910} = n\)

Round up to find the least number of days needed

Since we can't have a fraction of a day, we need to round up to the nearest whole number to find the least number of days needed to complete the journey. \(n = \lceil \sqrt{5910} \rceil = 77\) So, it takes the man 77 days to complete his journey.

Key concepts

Arithmetic Series

Arithmetic series are sequences of numbers in which each term is derived by adding a constant, called the common difference, to the previous term. For example, if you start with 1.5 and keep adding 0.1667, you would get 1.5, 1.6667, 1.8334, and so on. This pattern builds an arithmetic progression.
To understand or even construct an arithmetic series properly, you need:

• A starting number, also known as the first term \(a_1\).
• A common difference \(d\), which is the fixed amount added to each term.

These two components help you determine each following number in your sequence and the sum of the series over multiple terms.

Sum of Arithmetic Series

The sum of an arithmetic series is calculated using a specific formula that makes it easy to add up all the numbers in the series efficiently. The formula is:

• \(S_n = \frac{n}{2}(a_1 + a_n)\)

Here, \(S_n\) represents the total sum of the first \(n\) numbers, \(a_1\) is the first term, and \(a_n\) is the last term. This formula is incredibly valuable as it provides a quick way to find the sum without the need to add each individual term manually.
For example, if you want to find how far someone traveled over multiple days by progressively adding a certain distance, this formula simplifies the process of totaling those daily travels.

Arithmetic Progression Formula

The arithmetic progression formula focuses on finding any specific term in the sequence rather than the entire sum. This formula is:

• \(a_n = a_1 + (n-1) \cdot d\)

Here, \(a_n\) is the desired term, \(a_1\) is the initial term, \(d\) stands for the common difference, and \(n\) is the position of the term in the sequence.
This allows you to calculate, for example, the exact distance traveled on any specific day. Knowing \(a_1\) and \(d\) from our problem of the journey, you can plug those numbers into the formula to find out how much was traveled on the last day, \(a_n\), just before the journey completes.