Question

ms. clark earned $29,600 in her third year as a banker and $32,600 in her seventh year. how much did she make in her tenth year? assume that her yearly salaries form an arithmetic sequence.
○ $28,100
○ $34,850
○ $55,550
○ $58,100

Explanation:

Step1: Recall Arithmetic Sequence Formula

The formula for the \(n\)-th term of an arithmetic sequence is \(a_n = a_1+(n - 1)d\), where \(a_n\) is the \(n\)-th term, \(a_1\) is the first term, \(n\) is the term number, and \(d\) is the common difference.

We know that \(a_3=29600\) and \(a_7 = 32600\). Using the formula:
For \(n = 3\): \(a_3=a_1+(3 - 1)d=a_1 + 2d=29600\)
For \(n = 7\): \(a_7=a_1+(7 - 1)d=a_1+6d = 32600\)

Step2: Solve for Common Difference \(d\)

Subtract the first equation from the second equation:
\((a_1 + 6d)-(a_1 + 2d)=32600 - 29600\)
Simplify: \(4d=3000\), so \(d=\frac{3000}{4}=750\)

Step3: Solve for First Term \(a_1\)

Substitute \(d = 750\) into \(a_1+2d=29600\):
\(a_1+2\times750=29600\)
\(a_1+1500 = 29600\)
\(a_1=29600 - 1500=28100\)

Step4: Find the Tenth Term \(a_{10}\)

Use the formula \(a_n=a_1+(n - 1)d\) with \(n = 10\), \(a_1=28100\), \(d = 750\):
\(a_{10}=28100+(10 - 1)\times750\)
\(a_{10}=28100+9\times750\)
\(a_{10}=28100 + 6750=34850\)

Answer:

\(\$34,850\) (corresponding to the option \(\$34,850\))