CPM Homework Help : CCA Problem 11-47

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11-47.

Find an equation for each sequence.

n
t(n)
$1$
$7$
$2$
$4$
$3$
$1$
Step 1(a):
To find the equation, you need to know the initial term, when $n = 0$ .
Step 2(a):
When $n = 1$ , $t(n) = 7$ , and when $n = 2$ , $t(n) = 4$ , so you know that it decreases by $3$ from one term to the next.
Step 3(a):
To find the number before $7$ , you add $3$ , because $3$ was taken away to get $7$ . So the initial term is $10$ .
Step 4(a):
To get the number after $10$ , you take away $3$ once. To get the next number, you take away $3$ again. To get the next number you take away $3$ again. Since you are always subtracting $3$ , you multiply $3$ by $n$ to represent how many times you have subtracted $3$ . Now write your equation.
Answer (a):
$t(n)=7−3(n−1)$
$t(n)=10−3n$

Year
Cost
$2$
$\$2000$
$3$
$\$6000$
$4$
$\$18000$
Step 1(b):
This sequence is different than the one in part (a). You multiply instead of subtract.
Step 2(b):
The cost was $$2000$ in Year $2$ , but you need to know how much it was in Year $0$ to write the equation. Each year it is multiplied by $3$ . So the cost in Year $2$ is $3$ times what it was in Year $1$ , which is $3$ times what it was in Year $0$ . So if you divide the cost in Year $2$ by $9$ , you will have the cost from Year $0$ .
Step 3(b):
Similar to part (a), because you are always multiplying by $3$ , you can use a variable as the exponent of $3$ to represent how many times you have multiplied by $3$ . Now write your equation.
Answer (b):
$\textit{t}(\textit{n})=\left(\frac{2000}{3}\right)\cdot3^{\textit{n}-1}$
$t(n)=\left(\frac{2000}{9}\right)\cdot3^n$

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