Question 1027240: the first three terms of a geometric progression are 100,90 and 81.
Find out the common ratio?
find the sum to infinity of its terms?
can I be explained step by step how to figure this out
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! a geometric progression has the form An = A1 * r^(n-1)
An is the nth term.
A1 is the first term.
n is the number of terms.
your problem has the sequence 100, 90, 81, .....
the common ratio is .9 because 100 * .9 = 90 and 90 * .9 = 81.
the formula of An = A1 * r^(n-1) therefore becomes An = 100 * .9^(n-1)
for example, the third term, which you alreay know is .81, would be calculated using this forula as A3 = 100 * .9^(2.
you would then get A3 = 81.
this agrees with what you already know so the formula is confirmed as good.
you want to know the sum of this progression as n approaches infinity.
the formula for the sum of the geometric progression as n approaches infinity is:
Sum of geometric progression as n approaches infinity = A1 * 1 / (1-r)
for your progression, that would be 100 * 1/(1-.9) = 100 / .1 = 1000
the following reference discusses geometric progressions.
http://www.purplemath.com/modules/series5.htm
i used excel to analyze whether this was true or not.
it turns out to be true.
after about the 160th term, the sum reached 1000 and didn't go over.
the formula for the sum of the geometric progression to the nth term is:
sum to the nth term = A1 * (1 - r^n) / (1-r)
for the 160th term, that formula would become sum = 100 * (1 - .9^160) / (1-.9) which becomes sum = 100 * .9999999523/.10 which becomes equal to 999.9999523.
that's very close to 1000.
it gets closer as n gets higher, but never goes over 1000.
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