SOLUTION: Find all the geometric sequences such that the sum of the first two terms is 2 and the sum of the first three terms is 3. I got this: a + ar = 2 a + a(+- 1/√a) = 2 I

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Find all the geometric sequences such that the sum of the first two terms is 2 and the sum of the first three terms is 3. I got this: a + ar = 2 a + a(+- 1/√a) = 2 I       Log On


   

Question 175764: Find all the geometric sequences such that the sum of the first two terms is 2 and the sum of the first three terms is 3.
I got this:
a + ar = 2
a + a(+- 1/√a) = 2
I don't know what to do now. Please help!
Found 2 solutions by Edwin McCravy, gonzo:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Find all the geometric sequences such that the sum of the first two terms is 2 and the sum of the first three terms is 3. 

You have this system of equations:

system%28a%5B1%5D+%2B+a%5B2%5D+=+2%2C%0D%0Aa%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+=+3%29

Now if you subtract the first equations
term by term from the second equation
you get

a%5B3%5D+=+1

a%5B1%5Dr%5E%283-1%29+=+1

a%5B1%5Dr%5E2+=+1

a%5B1%5D+=+1%2Fr%5E2

a%5B2%5D+=+a%5B1%5Dr+=+%281%2Fr%5E2%29r+=+%281%2Fr%5Ecross%282%29%29cross%28r%29+=+1%2Fr

So the first three terms are

a%5B1%5D+=+1%2Fr%5E2
a%5B2%5D+=+1%2Fr
a%5B3%5D+=+1

and a%5Bn%5D is given by:
 
a%5Bn%5D+=+a%5B1%5Dr%5E%28n-1%29+=+%281%2Fr%5E2%29%28r%5E%28n-1%29%29+=+r%5E%28n-3%29

So all such geometric sequences are of the form:

1%2Fr%5E2, 1%2Fr, 1, r, r%5E2, r%5E3,···,r%5E%28n-3%29,···

Edwin

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
i forget the formulas all the time so i keep having to look them up.
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here's a website that contains geometric series formulas.
http://www.purplemath.com/modules/series5.htm
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there are others.
all you have to do is go to yahoo or google and search on geometric series, or geometric sequence, or sum of geometric sequence, etc.
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after you search a number of times, you'll get the hang of what keywords you should use.
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you can even look in your book.
since i don't have a book i use the web.
the web is useful because you get different ways of explaining the material and you can choose the way which is easiest for you to understand.
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my search led me to the following formula.
sum%28+a%5Bi%5D%2C+i=1%2C+n+%29+=+a%5B1%5D%2A%28%281-r%5En%29%2F%281-r%29%29
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my search also led me to the following formula for the nth term of a geometric series.
a%5Bn%5D+=+a%5B1%5D%2Ar%5E%28n-1%29
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you are given that the sum of the first 2 terms is equal to 2.
this means that:
a + ar = 2
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you are also given that the sum of the first 3 terms is equal to 3.
this means that:
a + ar + ar^2 = 3
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if you subtract the sum of the first 2 terms from the first 3 terms, you get the 3d term.
this means that:
a + ar + ar^2 - a - ar = 3 - 2
which results in:
ar^2 = 1
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if ar^2 = 1, then
a = (1/r^2)
or
r^2 = (1/a)
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i tried using r^2 = (1/a) but got nowhere.
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i then tried using a = (1/r^2) and got somewhere as follows:
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since the sum of the first 2 terms equals 2, this means that:
a + ar = 2
since a = (1/r^2), this also means that:
(1/r^2) + (1/r^2)*r = 2
which means that:
(1/r^2) + 1/r = 2, since r/r^2 = 1/r
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i then multiplied both sides of this equation by r^2 to remove the fractions.
doing this i got:
1 + r = 2r^2
subtracting (1+r) from both sides of the equation, i then got:
2r^2 - r - 1 = 0
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this factors out to be:
(2r+1) * (r-1) = 0
which yields the following results:
r = 1
or
r = -1/2
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either one or both of these values should help solve the equation.
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i tried r = 1 first.
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i already figured out that ar^2 = 1
if r = 1, this means that:
a*1^2 = 1
which means that:
a = 1.
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if a = 1, and the sum of the first terms is equal to 2, then the 2d term in the sequence must be equal to 1.
the sum of the first 3 terms is equal to 3, and the 3d term is equal to 1, so this holds up because 1 + 1 + 1 = 3.
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r = 1 is good.
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next i looked at r = -1/2 to see if this is also valid.
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since the 3d terms is equal to 1 and is also equal to ar^2, i solved for:
ar^2 = 1, using r = (-1/2)
this became:
a*(-1/2)^2 = 1
which became:
a*(1/4) = 1
which became:
a = 4
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if a = 4, and r = -1/2, then the second term would be:
4*(-1/2) = -2.
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since the first term is 4 and the second term is -2, then the sum of the first 2 terms is 2 which is valid.
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i then took the second term and multiplied it to get the 3d term which should equal 1.
-2 * (-1/2) = 1 which is equal to the third term.
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since the sum of the first 3 terms is 3, then:
4 - 2 + 1 should equal 3.
it does.
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your answer is:
r can either be 1 or (-1/2).
if r is 1, then a is 1.
if r is (-1/2), then a is 4
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what happens if you go out one further?
take n = 4
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if a = 1 and r = 1 this would be equal to:
1*1^3 = 1
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if a = 4 and r = (-1/2) this would be equal to:
1*(-1/2)^3 = -.5
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the fourth terms are not equal, and since the sum of the first 3 terms are equal, it follows that the sum of the first 4 terms will not be equal either.
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in order words, the sum of each of these geometric sequences are only equal when n = 2 or n = 3 and not after, or at least not every time after, and certainly not equal when n = 1.
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since the problem only states that the sum of the first 2 terms and the sum of the first 3 terms are equal, both answers are still valid because they satisfy the given criteria.
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