### Home > CCAA8 > Chapter 11 Unit 12 > Lesson CCA: 11.2.2 > Problem11-47

11-47.

Find an equation for each sequence.

1.  $n$ $t(n)$ $1$ $7$ $2$ $4$ $3$ $1$

To find the equation, you need to know the initial term, when $n = 0$.

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.

To find the number before $7$, you add $3$, because $3$ was taken away to get $7$. So the initial term is $10$.

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.

$t(n)=7−3(n−1)$
$t(n)=10−3n$

1.  Year Cost $2$ $\2000$ $3$ $\6000$ $4$ $\18000$

This sequence is different than the one in part (a). You multiply instead of subtract.

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 $0$.

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.

$\textit{t}(\textit{n})=\left(\frac{2000}{3}\right)\cdot3^{\textit{n}-1}$
$t(n)=\left(\frac{2000}{9}\right)\cdot3^n$