SOLUTION: find the sum of the first n terms of the G.P 5+15+45+......... what is the smallest number of terms that will give a total of 1000.

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Question 1083649: find the sum of the first n terms of the G.P 5+15+45+......... what is the smallest number of terms that will give a total of 1000.
Answer by KMST(5328) About Me  (Show Source):
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The first term is b%5B1%5D=5 .
The common ratio is r=15÷5=3 .
Term number n is b%5Bn%5D=5%2A3%5E%28n-1%29 .
The sum of the first n terms is
b%5B1%5D%28r%5En-1%29%2F%28r-1%29=5%283%5En-1%29%2F%283-1%29=5.%283%5En-1%29%2F2
If we want to find what is the smallest n
that will make the,sun of the first n terms no less than 100, our inequality is
5%283%5En-1%29%2F2%3E=1000
3%5En-1%3E=1000%2A2%2F5
3%5En-1%3E=400
3%5En%3E=401
3%5E6=729
3%5E5=243
The sum of first n terms that is no less than 1000 is the sum of the first 6 terms:
5+15+45+135+405+1215.
The sum of some terms cannot be exactly 1000.
For the sum of some terms of the form b%5Bk%2B1%5D=5%2A3%5Ek
(with k%3E=0 being some integer)
to be exactly 1000,
the sum of all the 3%5Ek factors has to be 200.´
All those factors are multiples of 3, except 3%5E0=1.
and the sum of multiples of 3 cannot be either 200 or 199.