SOLUTION: Two employees started to work in a company: The first started with a lump sum (fixed) salary in an amount of $6,000, the other started with a salary beginning at $4,800 at the firs

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Question 934966: Two employees started to work in a company: The first started with a lump sum (fixed) salary in an amount of $6,000, the other started with a salary beginning at $4,800 at the first year and increases by $100 in each of the following years.
1. After how many years will the two salaries be equal?
2. After how many years will the total amount received by the first employee equal to the total amount received by the second employee?
Answer by vicgonzerx(31) (Show Source):
You can put this solution on YOUR website!
FROM ARITHMETIC PROGRESSIONS
L=a+(n-1)d
L=last term
a=first term
n=number of terms
d=common difference
solve number 1. after how many years will the 2 salaries be equal
L=6000
a=4800
d=100
L=a+(n-1)d------formula a
6000=4800+(n-1)100
1200=100n-100
100n=1300
n=13 years .........the answer
for number 2 ..................
from thr formula S=((a+L)/2)n----formula b
S=sum of all terms
S=6000n
then substitue the formula a to formula b to get S=n/2[2a+(n-1)d]
6000n=n/2[2(4800)+(n-1)100]
12000n=n(9600+100n-100)
100n^2-2500nn=0
divide it all by 100 to make it -----n^2-25n=o
transpose n^2=25n
divide it all by n to make it n=25
the answer is 25 years
checking 6000(25)=25/2[9600+100(25)-100]
150000=150000--------------qed
try using quadratic equation to get 25 years