Asked by Anonymous
So this is what Im basically doing, I got how to do Number 1 and 2. This is the way I did it.
Im supposed to find the common ration and the sum number.
1. 4 + 2 + 1 + ...
The Common ratio is 0.5, so in decimal form it would be 1/2. I sort of get confused like this, I am never able to get the fraction form first, its always the decimal first, could someone tell me how to find the fraction first?
Anyways, this is what I did...
S = a/1-r
S = 4/1-0.5
S = 4/0.5 = 8
How would I do this the fraction way, I want to get used to doing it in fraction form, that would make things much easier.
2. 5 - 1 + 1/5 -...
Common ratio is -0.2, so that would be -1/5.
S = a/1-r
S = 5/1-(-0.2)
S = 5/1.2 = 4.166 = 25/6
I don't get how to do number 3, cause when I came up with a sum, I checked for the real answer, and it said, that there cant be any sum :\
-4 + 6 - 9 + ...
The common ratio would be -1.5
S = a/1-r
S = -4/1-(1.5)
S = -4/2.5
S = 1.6
I got that as an answer, so why can there be no sum?
Im supposed to find the common ration and the sum number.
1. 4 + 2 + 1 + ...
The Common ratio is 0.5, so in decimal form it would be 1/2. I sort of get confused like this, I am never able to get the fraction form first, its always the decimal first, could someone tell me how to find the fraction first?
Anyways, this is what I did...
S = a/1-r
S = 4/1-0.5
S = 4/0.5 = 8
How would I do this the fraction way, I want to get used to doing it in fraction form, that would make things much easier.
2. 5 - 1 + 1/5 -...
Common ratio is -0.2, so that would be -1/5.
S = a/1-r
S = 5/1-(-0.2)
S = 5/1.2 = 4.166 = 25/6
I don't get how to do number 3, cause when I came up with a sum, I checked for the real answer, and it said, that there cant be any sum :\
-4 + 6 - 9 + ...
The common ratio would be -1.5
S = a/1-r
S = -4/1-(1.5)
S = -4/2.5
S = 1.6
I got that as an answer, so why can there be no sum?
Answers
Answered by
bobpursley
You have missed some basic material.
if r is between -1 and 1, the series converges, and you can get an infinite series sum
If r is greater than 1, or less than -1, the series gets larger and larger, and diverges, and you cannot get a sum.
Consider the series as
10+100+1000+10000+... r=10. THe series diverges, no sum
if r is between -1 and 1, the series converges, and you can get an infinite series sum
If r is greater than 1, or less than -1, the series gets larger and larger, and diverges, and you cannot get a sum.
Consider the series as
10+100+1000+10000+... r=10. THe series diverges, no sum
Answered by
Anonymous
Ok, so I would just keep the question unanswered? How would I put the answer as?
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