# A Geometric progression problem.

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Here is my way of working out the solution, i would appreciate some other methods to give me some ideas of how best to do these kind of questions:

Q: the sum of the first 3 terms in a GP is 9 times less than the sum of the first 6 terms, find the ratio.

Here is my solution:

since: 9(a+ar+ar^2)= a+ar+ar^2+ar^3+ar^4+ar^5

8+8r+8r^2= r^3+r^4+r^5

factorising: 8(r^2+r+1)=r^3(1+r+r^2)

canceling: 8= r^3

therefore r=2.

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Here is my way of working out the solution, i would appreciate some other methods to give me some ideas of how best to do these kind of questions:
It looks right to me. I would have worked out this problem in the same manner, since it seems to call for an algebraic approach.

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Here is my way of working out the solution' date=' i would appreciate some other methods to give me some ideas of how best to do these kind of questions:

Q: the sum of the first 3 terms in a GP is 9 times less than the sum of the first 6 terms, find the ratio.

Here is my solution:

since: 9(a+ar+ar^2)= a+ar+ar^2+ar^3+ar^4+ar^5

8+8r+8r^2= r^3+r^4+r^5

factorising: 8(r^2+r+1)=r^3(1+r+r^2)

canceling: 8= r^3

therefore r=2.[/quote']

An arithmetic progression is, if i remember correctly, a sequence of numbers that are generated according to the rule that each successive term in the sequence is obtained from the previous one by the addition of a single number called the common difference.

By contrast, a geometric progression is, if i remember correctly, a sequence of numbers that are generated according to the rule that each successive term in the sequence is obtained from the previous one by multiplication of a single number, called the common ratio.

Ok so...

You have r as the common ratio.

And the first term is represented by a.

A(1) = a

A(2) = ar

A(3) = ar2

A(4) = ar3

A(5) = ar4

A(6) = ar5

The condition of your problem is that the sum of the first three terms is nine times less than the sum of the first six terms. In other words, 9 times the sum of the first three terms, is equal to the sum of the first six terms. That is:

9(a+ar+ar2) = (a+ar+ar2+ar3+ar4+ar5)

From here you just use algebra to solve for r, doesn't matter what order you perform your operations in, since there is one and only one answer.

First divide both sides by a, since a is necessarily nonzero.

If a=0, then there would be an infinite number of answers to the question, rather than a unique solution.

9(1+r+r2) = (1+r+r2+r3+r4+r5)

So...

8+8r+8r2 = r3+r4+r5

Exactly as you have...

8(1+r+r2) = r3 (1+r+r2)

The equation can be true if (1+r+r2) =0.

In the case where (1+r+r2) is not equal to zero, you can divide by it to obtain:

8 = r3=r*r*r

Whence it follows that r is equal to the cubed root of 8, which is 2.

As for the case where

1+r+r2 =0

You can use the quadratic formula to solve for r, and see what happens.

root1 = -1+ (1-4)1/2

root2 = -1- (1-4)1/2

As you can see, the discriminant is negative, which means the two roots above are complex numbers, not real numbers.

Hence, given that a is nonzero, and r is an element of the set of real numbers, there is one and only one answer to the question, and that is that r=2.

Regards

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Why would you try and say that (r^2+r+1) is equal to zero? That would invalidate the whole expression ie 8 . 0 = r^3 . 0 which would be 0=0, we get nowhere. Also if a=0 then we get 0=0 again. Since the 2 expressions that cancel are identities i didnt think it would matter what value they have, as they will always cancel to leave r^3=8.

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Why would you try and say that (r^2+r+1) is equal to zero? That would invalidate the whole expression ie 8 . 0 = r^3 . 0 which would be 0=0, we get nowhere. Also if a=0 then we get 0=0 again. Since the 2 expressions that cancel are identities i didnt think it would matter what value they have, as they will always cancel to leave r^3=8.

It is an issue with complex numbers. Let me go back and show you why.

8(1+r+r2) = r3 (1+r+r2)

The equation can be true if (1+r+r2) =0.

In the case where (1+r+r2) is not equal to zero' date=' you can divide by it to obtain:

8 = r[sup']3[/sup]=r*r*r

Whence it follows that r is equal to the cubed root of 8, which is 2.

As for the case where

1+r+r2 =0

You can use the quadratic formula to solve for r, and see what happens.

root1 = -1+ (1-4)1/2

root2 = -1- (1-4)1/2

As you can see, the discriminant is negative, which means the two roots above are complex numbers, not real numbers.

Now, some have a problem with the complex numbers, and others do not, but regardless, it has been highly developed over the past few centuries.

So here is what i did, during the solution of the problem.

First, I started with the information given, as true.

9(a+ar+ar2) = (a+ar+ar2+ar3+ar4+ar5[/sup'])

You did as well:

Q: the sum of the first 3 terms in a GP is 9 times less than the sum of the first 6 terms' date=' find the ratio.

Here is my solution:

since: 9(a+ar+ar^2)= a+ar+ar^2+ar^3+ar^4+ar^5[/quote']

Then, i had to recall the definition of a GP, which i already knew. Again, and this is from memory, a geometric progression is a sequence of numbers, where the next term in the sequence is obtained from the previous one, by multiplication of a fixed number r, called the common ratio. Now, you chose 'a' to denote the first term of the sequence, so, so did i.

Now, there was no mention about whether or not a=0, or r=0, or both. Just not enough information given. So I just decided to give an answer which covered all possibilities.

(0,0,0,0,0,0...) is a sequence I believe, according to the definition of a sequence. I'll check right now to see, but I already do think it is. And if not, I certainly could adjust the definition of sequence, to allow it to be. Hold on, let me grab a link.

Ok, I looked and didn't find a definition, but I know there is one in my Real analysis text. But the basic idea behind one can be seen by considering the sequence of odd numbers.

Starting with 1, we have:

(1,3,5,7,9,11,13,...)

And we can put that sequence in 1-1 correspondence with the sequence of natural numbers:

(1,2,3,4,5,6,7,...)

Let A(n) denote an arbitrary term of the sequence: (1,3,5,7,9,11,13,...)

ane let n denote an arbitrary term of the sequence: (1,2,3,4,5,6,7,...)

The two are related by: A(n) = 2n-1

So the basic idea, is that the elements of the sequence are ordered, a first, a second, a third, and so on.

Now, in the example above, consider Dn.... "change in n."

If you study, or have studied the finite discrete difference calculus, you will have come across the difference operator, defined below:

Df(x) = f(x+h) - f(x)

In the notation above, h is called the step size. In the sequence (1,2,3,4,5,...) the step size happens to be one, but this does no have to be the case.

Also, the difference operator, defined above, is used to define the derivative in the differential calculus.

So the point I am currently trying to make, is that all that is important in the notion of "sequence" is that the terms come in an order, you know like how things are ordered in time. Nothing more than this.

So, from that, (0,0,0,0,...) is a sequence.

Hence, if a=0, and not(r=0), then the following sequence is a geometric progression, with common ratio r:

(0,0r,0rr,0rrr,0rrrr,...)

which is equivalent to

(0,0,0,0,0,...)

Also, possible is this case:

not(a=0) and r=0

(a,a0,a00,a000,...)

which is equivalent to

(0,0,0,0,...)

In mathematics, it is always important to know what definitions you are using, so that you can explain yourself, at the very least.

So, like i said, i believe (0,0,0,0,0,..) is a sequence, so i don't think you can just ignore the two cases I just showed above, without more information given in the problem.

So I just handled all the possible cases, which leads to not just dividing by 1+r+rr, because I did not assume that it isn't zero.

In other words when I got to the following statement:

8(1+r+r2) = r3 (1+r+r2)

The statement above was certainly true, regardless of what r could be equal to.

All is fine up to there, and I was certain that all was fine.

But, there is a logical reason why you cannot divide by zero.

If you logically analyze the field axioms of the algebraic operations, you see that if you are using certain field axioms together, that permitting yourself to divide by zero, allows you to reach the conclusion that 0=1 and not(0=1).

And I am freely using the field axioms, so i have to be careful whenever I divide by something with a letter in it.

So here is what the logic is then...

8(1+r+r2) = r3 (1+r+r2)

Statement above is true for certain.

If (1+r+r2)=0, then RHS=0 and LHS =0, which makes statement true, as we know it is. Yet if that is the case, then we cannot divide by it, we have to leave things as they are.

At this point, I didn't know anything about the roots, i simply used the quadratic formula to have a look at them.

The quadratic formula can be obtained by completing the square, you learn how to do that, before you memorize the quadratic formula, because the one is used to deduce the other.

It's been years since I've actually derived the quadratic formula by completing the square, but i still remember how.

So, using the quadratic formula I found this:

You can use the quadratic formula to solve for r' date=' and see what happens.

root1 = -1+ (1-4)[sup']1/2

root2 = -1- (1-4)1/2

As you can see, the discriminant is negative, which means the two roots above are complex numbers, not real numbers.

Up till then, i didn't already know that the roots of that polynomial, were necessarily complex. But then at this point i did.

I looked back at your question, and didn't see any reason why r couldn't be a complex number.

So that finished off the case where (1+r+r2)=0.

Then I covered the mutally exclusive case, where the polynomial above is not equal to zero. In this case you can divide through it, and then you can rapidly see that r=2.

The two cases above are mutually exclusive, and collectively exhaustive, so i was done.

To summarize: Unless there is a reason why (0,0,0,0,0..) cannot be a sequence, or unless there is a reason why r cannot be a complex number, I don't see anything wrong with a comprehensive answer.

There was nothing at all wrong with what you did.

You just asked to see alternative approaches to answering the question, so i tried to see how I would solve it. I happen to like the geometric series, which is where i encountered arithmetic/geometric progressions in the first place, and this was many years ago.

Kind regards

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I follow your reasoning to a certain extent although i have not studied complex numbers or anything of that kind yet so it will probably make more sense once i have. I had assumed that because the first 6 terms were more than the first 3 that there could not be a=0 or r=0 because then we would still have a series but it would not increase as the terms progressed.

I appreciate your detailed explanations all the same.

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