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    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.
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    1) Multiply top and bottom by rt 2
    i.e.  \displaystyle \frac{a}{1-\frac{1}{\sqrt{2}}} . \frac{\sqrt 2}{\sqrt 2} = \frac{a\sqrt 2 }{\sqrt 2 - 1}

    Proceed to rationalise the denominator as usual.
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    (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)
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    1) Just multiply top and bottom of the equation by sqrt2 to get

    \frac{a\sqrt{2}}{\sqrt{2} - 1} - 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 .
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    (Original post by Swayum)
    1) Just multiply top and bottom of the equation by sqrt2 to get

    \frac{a\sqrt{2}}{\sqrt{2} - 1} - 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 .
    :devil:

    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
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    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.
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    (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.
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    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 ?
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    (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.
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    So I get:

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

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

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

    Thanks for your help.
 
 
 
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