word problem

tsh44

Junior Member
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Sep 4, 2005
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Hi this word problem says suppsoe a dotor earns 40000 dollars a year during his 1st year. THen it says suppose that each succeeding year the salary increases 10%. What is the total of his salary over the 10 years and how many years must he work to exceed a million dollars. How would I do this?
 
Hint: Consider geometric series, and look at using the series-sum formula.

Eliz.
 
What is the formula and how would I represent the 10% increase eahc year?
 
Not clear to me; are you looking for his salary during 10th year?
If so, use formula F = P(1 + r)^n
F = Future value (?)
P = Present value (40000)
r = rate of increase (.10)
n = number of years (10)

To get the year in which he'll hit a million:
40000(1 + .10)^n = 1000000
solve for n
 
tsh44 said:
What is the formula and how would I represent the 10% increase eahc year?

The formula for the sum of the first "n" terms of a geometric series where the first term is represented by a<SUB>1</SUB> and the common ratio is represented by r is

S<SUB>n</SUB> = (a<SUB>1</SUB> - a<SUB>1</SUB>*r<SUP>n</SUP>) / (1 - r)

The first year, the doctor earns $40,000.....this is the value of a<SUB>1</SUB>. We'd like the total he has earned in 10 years, so n = 10. Now, what about the value of r? Each year his salary increases by 10%, so he will earn 110% of the previous year's salary. For example, in his second year he will earn 110% of $40,000, which can be written as 40000*1.10. The following year, he'll earn 110% of (40000*1.10), or 40000*1.10*1.10, or 40000*1.10<SUP>2</SUP>. Do you see that the common ratio, then, is 1.10? That's the value you will use for r.

Substitute these values into the formula, and calculate his total earnings for 10 years.

To answer the second part of the question, substitute 1000000 for S<SUB>n</SUB> and solve the resulting equation for n.....
 
For the second part i got the equation to 1.1^n=3.5
How would I solve for n? Thanks for helping with the 1st part too.
 
tsh44 said:
For the second part i got the equation to 1.1^n=3.5
How would I solve for n? Thanks for helping with the 1st part too.
How did you get that?

40000(1.1)^n = 1000000
1.1^n = 1000000 / 40000 = 25
n = log(25) / log(1.1) = ~33.77 : so in 34th year
 
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