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# Questions on Geometric Sequences C2 watch

1. I just can't seem to get some of these questions on Geometric sequences:

1.) First term: a Common ratio: (1/square root 2). Find the sum to infinity.

I know this is just simplyfying surds but I can't understand what to do after:

a / 1-(1/sqrt 2).

2.) The sum of the infinite series 1 + r + (r^2) + (r+3)...... is K times larger than the sum of the series 1 - r + (r^2) - (r^3). Express r in terms of k.

So for this I have for the first series: a = 1, r = r

...so 1/1-r is the sum to infintiy. For the second one, what would be the common difference as it is alternating signs ?

Thanks for your time and help.
2. 1) Multiply top and bottom by rt 2
i.e.

Proceed to rationalise the denominator as usual.
3. (Original post by cyborg)
2.) The sum of the infinite series 1 + r + (r^2) + (r+3)...... is K times larger than the sum of the series 1 - r + (r^2) - (r^3). Express r in terms of k.

So for this I have for the first series: a = 1, r = r

...so 1/1-r is the sum to infintiy. For the second one, what would be the common difference as it is alternating signs ?

Thanks for your time and help.
Common ratio = (-r)
4. 1) Just multiply top and bottom of the equation by sqrt2 to get

- you should know how to rationalise this.

2) Is your second series 1 - r + r^2 - r^3 + r^4 + ...? The common ratio is then just -r.

*Edit*

Beaten .
5. (Original post by Swayum)
1) Just multiply top and bottom of the equation by sqrt2 to get

- you should know how to rationalise this.

2) Is your second series 1 - r + r^2 - r^3 + r^4 + ...? The common ratio is then just -r.

*Edit*

Beaten .

I've been up all night trying to do my maths, but I keep taking breaks to do easier maths on here! Should've finished my work several hours ago
6. For the first answer I got 2a + a(root 2) which is the right answer. Thanks for all the help.

For the second one, this is how far I've got:

Sum to infinity of the first series is a/(1-r) and for the second is a/(1+r). But now I need to express r in terms of k as the question says Sum of the first series is K times sum of the second.

So something like: a/(1-r) = ka/(1+r) ? Then how do I carry on ?

Thanks again.
7. (Original post by cyborg)
For the first answer I got 2a + a(root 2) which is the right answer. Thanks for all the help.

For the second one, this is how far I've got:

Sum to infinity of the first series is a/(1-r) and for the second is a/(1+r). But now I need to express r in terms of k as the question says Sum of the first series is K times sum of the second.

So something like: a/(1-r) = ka/(1+r) ? Then how do I carry on ?

Thanks again.
the k should be on the 1-r side.

multiply throughout the equation by (1+r)(1-r) and rearrange to get r on its own.
8. I still don't seem to get it.

If I multiply k / 1-r by (1+r)(1-r) I get k - kr^2 / 1-r ?
9. (Original post by cyborg)
I still don't seem to get it.

If I multiply k / 1-r by (1+r)(1-r) I get k - kr^2 / 1-r ?
You're not seeing the forest because of the trees!

the (1-r) terms cancel.
10. So I get:

k/(1-r) multiplied by (1-r)(1+r) so the (1-r) cancels giving me k+rk ?
11. I've got it

1+r = k-kr
r(k+1)=k-1

so r = (k-1)/(K+1) which is the correct answer.

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