Question
A computer programmer makes $60,000 in her first year of working at a company. She gets a 10 percent pay raise every year. Create a geometric series model for how much she makes in his first 5 years of working.(1 point)
Responses
∑5n=360,000(1.1)n−1
sigma-summation underscript n equals 3 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n minus 1 baseline
∑5n=160,000(0.1)n−1
sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 0.1 right parenthesis superscript n minus 1 baseline
∑5n=160,000(1.1)n−1
sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n minus 1 baseline
∑5n=560,000(1.1)n
Responses
∑5n=360,000(1.1)n−1
sigma-summation underscript n equals 3 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n minus 1 baseline
∑5n=160,000(0.1)n−1
sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 0.1 right parenthesis superscript n minus 1 baseline
∑5n=160,000(1.1)n−1
sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n minus 1 baseline
∑5n=560,000(1.1)n
Answers
GPT-4o mini
To model the programmer's salary over the first 5 years with a 10% raise each year, we can express her salary as follows:
- In Year 1, her salary is $60,000.
- In Year 2, her salary (after a 10% raise) is \( 60,000 \times 1.1 \).
- In Year 3, her salary is \( 60,000 \times (1.1)^2 \).
- In Year 4, her salary is \( 60,000 \times (1.1)^3 \).
- In Year 5, her salary is \( 60,000 \times (1.1)^4 \).
The total salary over these 5 years can be expressed as a geometric series:
\[
\text{Total Salary} = 60,000 + 60,000 \times 1.1 + 60,000 \times (1.1)^2 + 60,000 \times (1.1)^3 + 60,000 \times (1.1)^4
\]
This can be factored into the summation:
\[
\text{Total Salary} = \sum_{n=0}^{4} 60,000 \times (1.1)^n
\]
This is a geometric series with the first term \( a = 60,000 \) and common ratio \( r = 1.1 \).
Therefore, the appropriate expression from the provided options, where \( n \) starts from 1, would be:
\[
\sum_{n=1}^{5} 60,000 \times (1.1)^{n-1}
\]
The correct choice from the given responses is:
**sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n minus 1 baseline**
- In Year 1, her salary is $60,000.
- In Year 2, her salary (after a 10% raise) is \( 60,000 \times 1.1 \).
- In Year 3, her salary is \( 60,000 \times (1.1)^2 \).
- In Year 4, her salary is \( 60,000 \times (1.1)^3 \).
- In Year 5, her salary is \( 60,000 \times (1.1)^4 \).
The total salary over these 5 years can be expressed as a geometric series:
\[
\text{Total Salary} = 60,000 + 60,000 \times 1.1 + 60,000 \times (1.1)^2 + 60,000 \times (1.1)^3 + 60,000 \times (1.1)^4
\]
This can be factored into the summation:
\[
\text{Total Salary} = \sum_{n=0}^{4} 60,000 \times (1.1)^n
\]
This is a geometric series with the first term \( a = 60,000 \) and common ratio \( r = 1.1 \).
Therefore, the appropriate expression from the provided options, where \( n \) starts from 1, would be:
\[
\sum_{n=1}^{5} 60,000 \times (1.1)^{n-1}
\]
The correct choice from the given responses is:
**sigma-summation underscript n equals 1 overscript 5 End Scripts 60,000 left parenthesis 1.1 right parenthesis superscript n minus 1 baseline**