SOLUTION: A starting salary for a teacher is $25,000 and there is an annual increase of 3%. (a) how much will the teacher earn in their 10th year? (b) how much will the teacher earn i

Question 1034586: A starting salary for a teacher is $25,000 and there is an annual increase of 3%.
(a) how much will the teacher earn in their 10th year?
(b) how much will the teacher earn in total during a 35 year teaching career?
(c) find the first year in which the teacher earns more than $35,000?
(d) how many years would the teacher have to work in order to earn a total of $1 million?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
this looks like a geometric progression where the common ratio is 1.03.

the nth term of a geometric progression is given by the formula:

An = A1 * r^(n-1).

when A1 = 25,000, and r = 1.03, the formula becomes:

An = 25,000 * 1.03^(n-1).

when n = 1, you get A1 = 25,000 * 1.03^(0) = 25,000.
when n = 2, you get A2 = 25,000 * 1.03^1 = 25,750.

the sum of a geometric progression is given by the formula:

Sn = A1 * (1 - r^n) / (1-r)

when n = 1, you get S1 = 25,000 * (1 - 1.03^1) / (1-1.03) = 25,000
when n = 2, you get S2 = 25,000 * (1 - 1.03^2) / (1 - 1.03) = 50,750.

since we know what A1 and A2 are from above, we can just add them together to get 25,000 + 25,750 = 50,750.

since this is the same as the Sn formula when n = 2, we have just confirmed that the formulas are good.

now we can answer the questions you have.

the questions with their answers are shown below:

(a) how much will the teacher earn in their 10th year?

A10 = A1 * r^(n-1) becomes:
A10 = 25,000 * 1.03^9 which becomes:
A10 = 32,619.3296
the teacher will have earned 32,619.3296 by the end of their 10th year.

(b) how much will the teacher earn in total during a 35 year teaching career?

S35 = A1 * (1 - r^35) / (1-r) becomes:
S35 = 25,000 * (1 - 1.03^35) / (1-1.03) which becomes:
S35 = 1,511,552.045
the teacher will earn a total of 1,511,552.045 by the end of their 35th year.

(c) find the first year in which the teacher earns more than $35,000?

An = A1 * r^(n-1) becomes:
35,000 = 25,000 * 1.03^(n-1).
divide both sides of this equation by 25,000 to get:
35,000 / 25,000 = 1.03^(n-1)
take the log of both sides of this equation to get:
log(35,000/25,000) = log(1.03^(n-1)
since log(a^b) = b*log(a), this equation becomes:
log(35,000/25,000 + (n-1) * log(1.03)
divide both sides of this equation by log(1.03) to get:
log(35,000/25,000) / log(1.03) = n-1.
add 1 to both sides of this equation to get:
log(35,000/25,000) / log(1.03) + 1 = n
solve for n to get n = log(35,000/25,000) / log(1.03) + 1 = 12.38314854
confirm by replacing n in the equation of 35,000 = 25,000 * 1.03^(n-1) to get:
35,000 = 25,000 * 1.03^(12.38314854-1).
you will get 35,000 = 35,000.
when n = 12, the teacher earns 25,000 * 1.03^11 = 34,605.85
this means the teacher earns 34,605.85 by the end of the 12th year.
when n = 13, the teacher earns 25,000 * 1.03^(12 = 35,644.02.
this means the teacher earns 35,644.01 by the end of the 13th year.
this means that the first year that the teacher earns 35,000 is sometime between the end of the 12th year and the end of the 13th year.
this puts the first year that the teacher first earns 35,000 in the 13th year.

(d) how many years would the teacher have to work in order to earn a total of $1 million?

Sn = A1 * (1-r^n) / (1-r) becomes:
1,000,000 = 25,000 * (1-1.03^n) / (1-1.03)
divide both sides of this equation by 25,000 to get:
1,000,000/25,000 = (1-1.03^n) / (1-1.03)
multiply both sides of this equation by (1-1.03) to get:
(1,000,000/25,000) * (1-1.03) = 1 - 1.03^n
simplify to get:
-1.2 = 1 - 1.03^n.
subtract 1 from both sides of this equation to get:
-2.2 = -1.03^n
multiply both sides of this equation by -1 to get:
2.2 = 1.03^n
take the log of both sides of this equation to get:
log(2.2) = log(1.03^n)
since log(a^b) = b*log(a), this equation becomes:
log(2.2) = n * log(1.03)
divide both sides of this equaiton by log(1.03) to get:
log(2.2) / log(1.03) = n
solve for n to get n = log(2.2) / log(1.03) = 26.67419857.
replace n in the original equation of 1,000,000 = 25,000 * (1-1.03^n) / (1-1.03) to get:
1,000,000 = 25,000 * (1-1.03^26.67419857) / (1-1.03) = 1,000,000.
this confirms the solution is correct.
when n = 26, the teacher earns a total of S26 = 25,000 * (1 - 1.03^26)/(1-1.03) = 96,3826.06.
this means the teacher earns a total of 96,3826.06 by the end of the 26th year.
when n = 27, the teacher earns a total of S27 = 25,000 * (1-1.03^27)/(1-1.03) = 1,017,740.84.
this means the teacher earns a total of 1,017,740.84 by the end of the 27th year.
this means that the first year that the teacher earns 1,000,000 is sometime between the end of the 26th year and the end of the 27th year.
this puts the first year that the teacher first earns 1,000,000 in the 27th year.

RELATED QUESTIONS
In 2014, the median salary nationwide for a high school teacher was $56,310 per year. The (answered by ewatrrr)

Two people begin working on the same day for the same company. The starting salary of... (answered by josgarithmetic)

in a certain school system, the starting salary for a teacher is $12,000 with an annual... (answered by Mathtut)

Hello! I am confused with this word problem in which I have to use geometric/arithmetic... (answered by rfer)

In 2014, the median salary nationwide for a high school teacher was $56,310 per year. The (answered by Theo)

here is the question..... the starting salary for a first year teacher is 25700. each... (answered by stanbon)

A teacher starting salary of $40,000 would be how much after 30 years with a 4% increase... (answered by checkley79)

Please help me to do this problem. Thank you very much. Here it is: A clerk starts... (answered by jojo14344)