Question 1042512: Please help me solve this problem:
Dr. Jane is studying the growing population of a species of jellyfish in a body of water. Each year, she records the population of jellyfish. The first year, she recorded a population of 17 jellyfish. Every following year, the population increased by four times the previous year’s population. At this rate, what will be the jellyfish population in the 9th year of her study?
A. 1,485,477
B. 5,941,925
C. 371,371
D. 371,365
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i couldn't find any answer that was one of your selections.
in year 1, she has 17.
in year 2, she has 17 + 4 * 17 = 5 * 17 = 85
in year 3, she has 85 + 4 * 85 = 5 * 85 = 425
carry this out to year 9 and she has 17 * 5^8 = 6640625.
the key phrase in the problem statement is:
increased by four times the previous year’s population.
that means you take the current year's population and multiply it by 4 and then add it to the current year's population to get the next year's population.
if x in the first year, then it would be x + 4*x in the second year.
x + 4*x is equal to 5*x.
you wind up with a geometric progression where A9 = 17 * 5^8 = 6640625.
the general form of a geometric progression is An = A1 * r^(n-1).
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