Robin bought a computer for $$ 1, 250$ . It will depreciate, or decrease in value, by $10 %$ each year that she owns it.
a) Is the sequence formed by the value at the beginning of each year arithmetic, geometric, or neither? Explain.
b) Write an explicit formula to represent the sequence.
c) Find the value of the computer at the beginning of the $6^{t h} year$ .

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Solution

Step 1: Find the sequence formed by the value at the beginning of each year arithmetic, geometric, or neither.
Let us find the first three terms of the sequence.
Since the first term of the sequence is $a_{1} = 1250$ .
Since the value will depreciate, or decrease in value, by $10 %$ each year that she owns it
The second term of the sequence:
Subtract $10 %$ of the value from the original value $1250$ .
$\begin{array}{rcl} 1250 of 10 % & = & 1250 \times \frac{10}{100} \\ = & 125 \end{array}$
Subtract $125$ from the first term:
$1250 - 125 = 1125$
Find the third term of the sequence:
Subtract $10 %$ of the value from the second value $1125$ .
$\begin{array}{rcl} 1125 of 10 % & = & 1125 \times \frac{10}{100} \\ = & 112.5 \end{array}$
Subtract $125$ from the second term:
$1125 - 112.5 = 1012.5$
Therefore, the first three terms of the sequence will be $1250, 1125, and 1012.5$ .
The ratio $r$ to prove the sequence is arithmetic, geometric, or neither:
Substitute $a_{n - 1} = 1250 and a_{n} = 1125$ :
$\begin{array}{rcl} r & = & \frac{a_{n}}{a_{n - 1}} \\ \Rightarrow & r = \frac{1125}{1250} \\ \Rightarrow & r = 0.9 \end{array}$
Substitute $a_{n - 1} = 1125 and a_{n} = 1012.5$ :
$\begin{array}{rcl} r & = & \frac{a_{n}}{a_{n - 1}} \\ \Rightarrow & r = \frac{1012.5}{1125} \\ \Rightarrow & r = 0.9 \end{array}$
Recall the definition of a geometric sequence, which is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by, a fixed non-zero number called the common ratio.
Therefore, the sequence is formed by the geometric sequence.
Step 2: Write an explicit formula to represent the sequence.
The general formula for the geometric sequence is,
$a_{n} = a_{1} r^{n - 1}$
Substitute $a_{1} = 1250 and r = 0.9$ :
$\begin{array}{rcl} a_{n} & = & a_{1} r^{n - 1} \\ \Rightarrow & a_{n} = (1250) {(0.9)}^{n - 1} \end{array}$
Therefore, the required explicit formula is $\begin{array}{rcl} a_{n} & = & (1250) {(0.9)}^{n - 1} \end{array}$ .
Step 3: Find the value of the computer at the beginning of the $6 th$ year.
The explicit formula for the geometric sequence at $n = 6$ is,
$\begin{array}{rcl} a_{n} & = & (1250) {(0.9)}^{n - 1} \\ \Rightarrow & a_{6} = (1250) {(0.9)}^{6 - 1} \\ \Rightarrow & a_{6} = (1250) {(0.9)}^{5} \\ \Rightarrow & a_{6} = $ 738.1125 \end{array}$
Therefore, the value of the computer at the beginning of the $6^{t h}$ year will be $$ 738.1125$ .

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