Example 5
Completion requirements
Kale has ordered two packages online. The shipping was \($13.00\) for a package that weighed \(3\) kg and \($25.00\) for a package that weighed \(6\) kg. Assuming that the shipping costs follow an arithmetic sequence, determine the general formula for shipping costs based on package weight.
Step 1: Write down what is given.
In this problem, you are not given the first term or the common difference. Instead, you are given \(t_3 \) and \(t_6 \) . In order to solve the problem, you can set up a system of linear equations involving two equations to determine two variables.
For the package weighing 3 kg: Equation I
For the package weighing 6 kg: Equation II
Step 2: Set up a system, and solve for \(d\).
Taking these two equations, subtract Equation II from Equation I, which results in the elimination of \(t_1 \). Then, you can solve for \(d \).
You may notice that the two terms are three steps from each other; therefore, \(-3d = (13 - 25) \) or \( -3d = -12 \)
Step 3: Solve for \(t_1\).
Use one of the equations to find \(t_1 \).
\(\begin{align}
25 &= t_1 + 5d \\
25 &= t_1 + 5\left( 4 \right) \\
25 &= t_1 + 20 \\
5 &= t_1 \\
\end{align}\)
Step 4: Determine the general formula.
Substitute 5 for \(t_1 \) and 4 for \(d \) into the general term, the formula.
\(\begin{align}
t_n &= t_1 + \left( {n - 1} \right)d \\
t_n &= 5 + \left( {n - 1} \right)4 \\
t_n &= 5 + 4n - 4 \\
t_n &= 1 + 4n \\
\end{align}\)
Where \(t _n \) is the total shipping cost and \(n \) is the weight of the package, in kilograms.
In this problem, you are not given the first term or the common difference. Instead, you are given \(t_3 \) and \(t_6 \) . In order to solve the problem, you can set up a system of linear equations involving two equations to determine two variables.
For the package weighing 3 kg: Equation I
\(\begin{align}
t_3 &= 13 \\
t_1 &= ? \\
n &= 3 \\
d &= ? \\
\end{align}\)
t_3 &= 13 \\
t_1 &= ? \\
n &= 3 \\
d &= ? \\
\end{align}\)
\(\begin{align}
t_n &= t_1 + \left( {n - 1} \right)d \\
13 &= t_1 + \left( {3 - 1} \right)d \\
13 &= t_1 + 2d \\
\end{align}\)
t_n &= t_1 + \left( {n - 1} \right)d \\
13 &= t_1 + \left( {3 - 1} \right)d \\
13 &= t_1 + 2d \\
\end{align}\)
For the package weighing 6 kg: Equation II
\(\begin{align}
t_6 &= 25 \\
t_1 &= ? \\
n &= 6 \\
d &= ? \\
\end{align}\)
t_6 &= 25 \\
t_1 &= ? \\
n &= 6 \\
d &= ? \\
\end{align}\)
\(\begin{align}
t_n &= t_1 + \left( {n - 1} \right)d \\
25 &= t_1 + \left( {6 - 1} \right)d \\
25 &= t_1 + 5d \\
\end{align}\)
t_n &= t_1 + \left( {n - 1} \right)d \\
25 &= t_1 + \left( {6 - 1} \right)d \\
25 &= t_1 + 5d \\
\end{align}\)
Step 2: Set up a system, and solve for \(d\).
Taking these two equations, subtract Equation II from Equation I, which results in the elimination of \(t_1 \). Then, you can solve for \(d \).
\[\begin{align}
13 &= \cancel{t_1} + 2d \\
-(25 &= \cancel{t_1} + 5d) \\
\hline \\
-12 &= \qquad -3d \\
4 &= \qquad \enspace d \\
\end{align}\]
13 &= \cancel{t_1} + 2d \\
-(25 &= \cancel{t_1} + 5d) \\
\hline \\
-12 &= \qquad -3d \\
4 &= \qquad \enspace d \\
\end{align}\]
You may notice that the two terms are three steps from each other; therefore, \(-3d = (13 - 25) \) or \( -3d = -12 \)
Step 3: Solve for \(t_1\).
Use one of the equations to find \(t_1 \).
\(\begin{align}
25 &= t_1 + 5d \\
25 &= t_1 + 5\left( 4 \right) \\
25 &= t_1 + 20 \\
5 &= t_1 \\
\end{align}\)
Step 4: Determine the general formula.
Substitute 5 for \(t_1 \) and 4 for \(d \) into the general term, the formula.
\(\begin{align}
t_n &= t_1 + \left( {n - 1} \right)d \\
t_n &= 5 + \left( {n - 1} \right)4 \\
t_n &= 5 + 4n - 4 \\
t_n &= 1 + 4n \\
\end{align}\)
Where \(t _n \) is the total shipping cost and \(n \) is the weight of the package, in kilograms.
