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Eyad

  • 3 years ago

SUPER G.S QUESTION : ___________________ An infinite G.s the sum of the 1st n terms =b ,and the sum of the first 2n terms = c , and the sum of the first 3n terms = d , PROVE THAT (b,c-d,d-c,.....) is an infinite G.s and the sum of any number of its terms cant exceed the sum of the sum of the original sequence up to infinity ......

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  1. experimentX
    • 3 years ago
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    GS??

  2. Eyad
    • 3 years ago
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    Geometric sequence :D

  3. experimentX
    • 3 years ago
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    how that c-d is geometric mean of preceding and suceeding term

  4. experimentX
    • 3 years ago
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    show that (c-d)^2 = b * (d-c)

  5. Eyad
    • 3 years ago
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    @experimentX : he want to prove that these sequence is an infinite G.s ,not a Geometric sequence and since your working on the Geometric Mean that will lead you to the Geometric sequence ...

  6. Eyad
    • 3 years ago
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    these sequences are *

  7. experimentX
    • 3 years ago
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    do this first ... then use induction!!

  8. Eyad
    • 3 years ago
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    ok ,what about the next require ? i dont even understand what he want !_!

  9. experimentX
    • 3 years ago
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    \[ S_n = \frac{n(2a + (n-1)d)}{2}\] \[ S_{2n} = \frac{2n(2a + (2n-1)d)}{2}\] \[ S_{3n} = \frac{3n(2a + (3n-1)d)}{2}\] Find the differences, \[ S_{2n} - S_{n} = \frac{2n(2a + (2n-1)d)}{2} -\frac{n(2a + (n-1)d)}{2} = \frac{n}{2}(2a + (3n - 1)d) \] \[ S_{3n} - S_{2n} = \frac{3n(2a + (3n-1)d)}{2} -\frac{2n(2a + (2n-1)d)}{2} = \frac{n}{2}(2a + (5n - 1)d) \]

  10. experimentX
    • 3 years ago
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    looks like something went wrong ... verify that these are in Geometric progression first

  11. asnaseer
    • 3 years ago
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    experimentX: the question says "geometric sequence" not "arithmetic sequence", so the formula for the sum should be:\[S_n=\frac{a(1-r^n)}{1-r}\]

  12. asnaseer
    • 3 years ago
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    if you use this, then you should be able to do the question.

  13. experimentX
    • 3 years ago
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    Oh .. yes that's why i was getting this http://www.wolframalpha.com/input/?i=255*55+%3D+155^2

  14. Eyad
    • 3 years ago
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    yea thats right @asnaseer , anyways tysmm of your effort @experimentX

  15. asnaseer
    • 3 years ago
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    Also, Eyad, I think you may have a mistake in the question. I think it should say: "PROVE THAT (b, c-b, d-c, .....) is an ..." and not: "PROVE THAT (b, c-d, d-c, .....) is an ..."

  16. asnaseer
    • 3 years ago
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    note the 2nd term should be "c-b" not "c-d"

  17. asnaseer
    • 3 years ago
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    otherwise 3rd term = -(2nd term)

  18. asnaseer
    • 3 years ago
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    which doesn't seem to make sense

  19. asnaseer
    • 3 years ago
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    do you agree Eyad?

  20. Eyad
    • 3 years ago
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    I copied it exactly as its written from the book ,though i swear its doesn't make sense ,i totally agreeee

  21. asnaseer
    • 3 years ago
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    must be a misprint

  22. Eyad
    • 3 years ago
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    I think i'am gonna leave it and ask da prof. ,and if i got an answer i will surely share it to you guys ... Tysm for your efforts

  23. experimentX
    • 3 years ago
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    also something quite not right ... http://www.wolframalpha.com/input/?i=%282^10-1%29%28%282^30-1%29-%282^20-1%29%29+%3D+%282^20-1%29^2 I'm beginning to see things ... i think i should get some sleep .. lol

  24. asnaseer
    • 3 years ago
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    you should be able to use the formula I gave you to prove this fairly easily

  25. Eyad
    • 3 years ago
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    maybe ....

  26. asnaseer
    • 3 years ago
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    I can do the first step for you if you want?

  27. Eyad
    • 3 years ago
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    I dont want to bother you asnaseer i think iam gonna make sure first if the question is written correct then we shall begin again ,what do u think ?

  28. asnaseer
    • 3 years ago
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    I am 100% sure its a misprint - and it is no bother at all to help you - always a pleasure :)

  29. Eyad
    • 3 years ago
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    tysm @asnaseer ,ty too @experimentX ^_^ You guys rocks

  30. asnaseer
    • 3 years ago
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    ok, we know:\[S_n=\frac{a(1-r^n)}{1-r}=b\]therefore the sum of the first 2n terms must be:\[S_{2n}=\frac{a(1-r^{2n})}{1-r}=c\]similarly, the sum of the first 3n terms must be:\[S_{3n}=\frac{a(1-r^{3n})}{1-r}=d\]

  31. asnaseer
    • 3 years ago
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    therefore:\[c-b=\frac{a(1-r^{2n})}{1-r}-\frac{a(1-r^{n})}{1-r}=\frac{a(1-r^{2n}-1+r^n)}{1-r}=\frac{a(r^n-r^{2n})}{1-r}=\frac{ar^n(1-r^{n})}{1-r}\]\[\qquad=br^n\]

  32. Eyad
    • 3 years ago
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    yea...

  33. Eyad
    • 3 years ago
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    yes ,that right i have just proved it ..continue Please

  34. asnaseer
    • 3 years ago
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    sorry that should be - similarly, you should be able to show:\[d-c=br^{2n}\]

  35. asnaseer
    • 3 years ago
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    so you will be left with the sequence:\[b,c-b,d-c,...\]which equals:\[b,br^n,br^{2n},...\]

  36. asnaseer
    • 3 years ago
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    which is a geometric sequence

  37. Eyad
    • 3 years ago
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    aha ,now i will be able to prove it can be added to infinite G.s

  38. asnaseer
    • 3 years ago
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    yes

  39. Eyad
    • 3 years ago
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    ty ,ima be sure of the 2nd require and then call you back :DDD

  40. asnaseer
    • 3 years ago
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    ok - good luck! :)

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