Respuesta :
Answer is B. 34,850.
Using formula Un = a+(n-1)d
For third year,
29600 = a + 2d
For seventh year,
32600 = a + 6d
solve the simultaneous equations to find for a, and d, and substitute into U10 = a + 9d (from formula.) a should be 28100, and d should be 750.
Using formula Un = a+(n-1)d
For third year,
29600 = a + 2d
For seventh year,
32600 = a + 6d
solve the simultaneous equations to find for a, and d, and substitute into U10 = a + 9d (from formula.) a should be 28100, and d should be 750.
Assume it's an arithmetic sequence. a stands for the first salary, d stands for difference of each year, n stands the number of year
General formula for arithmetic sequence is
an = a + (n - 1)d
Given
a₃ = a + (3 - 1)d = 29,600
a₇ = a + (7 - 1)d = 32,600
Work on it:
First, with elimination, find the value of d
a + 2d = 29,600
a + 6d = 32,600
--------------------- - (substract)
-4d = -3,000
d = 750
The difference of each year is $750
Second, find the value of a with subtitution
a + 2d = 29,600
a + 2(750) = 29,600
a + 1,500 = 29,600
a = 29,600 - 1,500
a = 28,100
The salary on first year is $28,100
Third, find the salary in tenth year
an = a + (n-1)d
a₁₀ = a + (10 - 1)d
a₁₀ = 28,100 + 9(750)
a₁₀ = 28,100 + 6,750
a₁₀ = 34,850
Her salary in tenth year is $34,850. The answer is b
General formula for arithmetic sequence is
an = a + (n - 1)d
Given
a₃ = a + (3 - 1)d = 29,600
a₇ = a + (7 - 1)d = 32,600
Work on it:
First, with elimination, find the value of d
a + 2d = 29,600
a + 6d = 32,600
--------------------- - (substract)
-4d = -3,000
d = 750
The difference of each year is $750
Second, find the value of a with subtitution
a + 2d = 29,600
a + 2(750) = 29,600
a + 1,500 = 29,600
a = 29,600 - 1,500
a = 28,100
The salary on first year is $28,100
Third, find the salary in tenth year
an = a + (n-1)d
a₁₀ = a + (10 - 1)d
a₁₀ = 28,100 + 9(750)
a₁₀ = 28,100 + 6,750
a₁₀ = 34,850
Her salary in tenth year is $34,850. The answer is b