brizgney58z

2023-02-21

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 ${6}^{th}\mathrm{year}$.

Marquise Cole

Beginner2023-02-22Added 3 answers

Discover the order created by the arithmetic, geometric, or neither value at the start of each year.

Let's locate the sequence's first three terms.

Since the first term of the sequence is $a1=1250$.

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 $1250$.

$1250\mathrm{of}10\%=1250\times \frac{10}{100}=125$

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 $1125$.

$1125\mathrm{of}10\%=1125\times \frac{10}{100}=112.5$

Subtract $125$ from the second term:

$1125-112.5=1012.5$

Therefore, the first three terms of the sequence will be $1250,1125,\mathrm{and}1012.5$.

The ratio $r$ to prove the sequence is arithmetic, geometric, or neither:

Substitute $an-1=1250\mathrm{and}an=1125$:

$r=\frac{a}{n}\Rightarrow r=\frac{1125}{1250}\Rightarrow r=0.9$

Substitute $an-1=1125\mathrm{and}an=1012.5$:

$r=\frac{a}{n}\Rightarrow r=\frac{1012.5}{1125}\Rightarrow r=0.9$

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.

Write an explicit formula to represent the sequence.

The general formula for the geometric sequence is,

$an=a1{r}^{n-1}$

Substitute $a1=1250\mathrm{and}r=0.9$:

$an=a1{r}^{n-1}\Rightarrow an=\left(1250\right){\left(0.9\right)}^{n-1}$

Therefore, the required explicit formula is $an=\left(1250\right){\left(}^{0.9}$.

Find the value of the computer at the beginning of the $6\mathrm{th}$ year.

The explicit formula for the geometric sequence at $n=6$ is,

$an=\left(1250\right){(0.9)}^{n-1}\Rightarrow a6=\left(1250\right){(0.9)}^{6-1}\Rightarrow a6=\left(1250\right){(0.9)}^{5}\Rightarrow a6=\$738.1125$

Therefore, the value of the computer at the beginning of the ${6}^{th}$ year will be $\$738.1125$.

Let's locate the sequence's first three terms.

Since the first term of the sequence is $a1=1250$.

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 $1250$.

$1250\mathrm{of}10\%=1250\times \frac{10}{100}=125$

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 $1125$.

$1125\mathrm{of}10\%=1125\times \frac{10}{100}=112.5$

Subtract $125$ from the second term:

$1125-112.5=1012.5$

Therefore, the first three terms of the sequence will be $1250,1125,\mathrm{and}1012.5$.

The ratio $r$ to prove the sequence is arithmetic, geometric, or neither:

Substitute $an-1=1250\mathrm{and}an=1125$:

$r=\frac{a}{n}\Rightarrow r=\frac{1125}{1250}\Rightarrow r=0.9$

Substitute $an-1=1125\mathrm{and}an=1012.5$:

$r=\frac{a}{n}\Rightarrow r=\frac{1012.5}{1125}\Rightarrow r=0.9$

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.

Write an explicit formula to represent the sequence.

The general formula for the geometric sequence is,

$an=a1{r}^{n-1}$

Substitute $a1=1250\mathrm{and}r=0.9$:

$an=a1{r}^{n-1}\Rightarrow an=\left(1250\right){\left(0.9\right)}^{n-1}$

Therefore, the required explicit formula is $an=\left(1250\right){\left(}^{0.9}$.

Find the value of the computer at the beginning of the $6\mathrm{th}$ year.

The explicit formula for the geometric sequence at $n=6$ is,

$an=\left(1250\right){(0.9)}^{n-1}\Rightarrow a6=\left(1250\right){(0.9)}^{6-1}\Rightarrow a6=\left(1250\right){(0.9)}^{5}\Rightarrow a6=\$738.1125$

Therefore, the value of the computer at the beginning of the ${6}^{th}$ year will be $\$738.1125$.