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kryton1212

The general term of an infinite geometric sequence is given by: T(n)=6[(x+1)/8)^(n) (a) Find the range of the values of x for which the sum to infinity exists. (b) Find the sum to infinity of the sequence in terms of x. (c) If the sum to infinity of the sequence is 18, find the value of x.

  • 9 months ago
  • 9 months ago

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  1. kryton1212
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    \[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]

    • 9 months ago
  2. kryton1212
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    @hartnn @Callisto

    • 9 months ago
  3. hartnn
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    the sum to infinity will converge(exist) if \(|r|<1\) can you find 'r' from your general term ?

    • 9 months ago
  4. hartnn
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    general equation, \(T(n)=a_1r^{n-1}\)

    • 9 months ago
  5. kryton1212
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    umm..

    • 9 months ago
  6. hartnn
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    find a1 and 'r' both from your general term, you will need both

    • 9 months ago
  7. kryton1212
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    'r' is the ratio?

    • 9 months ago
  8. primeralph
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    |dw:1372921816382:dw|

    • 9 months ago
  9. kryton1212
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    ??

    • 9 months ago
  10. hartnn
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    compare, try to make the power as (n-1) \(\large 6 (\dfrac{x+1}{8})^n = 6 \dfrac{x+1}{8}(\dfrac{x+1}{8})^{n-1}\) now compare this with a1 r^{n-1} find a1 and r

    • 9 months ago
  11. hartnn
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    |dw:1372923030703:dw|

    • 9 months ago
  12. hartnn
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    then for the range of values of x, just do |r| <1

    • 9 months ago
  13. kryton1212
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    ahhh...

    • 9 months ago
  14. hartnn
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    @Jhannybean what! ?

    • 9 months ago
  15. primeralph
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    @Jhannybean Why?

    • 9 months ago
  16. hartnn
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    ohh..because @primeralph wrote "and" there! its actually "or"

    • 9 months ago
  17. primeralph
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    |dw:1372923436567:dw|

    • 9 months ago
  18. primeralph
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    @Jhannybean The x is only a part of r.

    • 9 months ago
  19. Jhannybean
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    Oh nevermind.... I was looking at it the wrong way.

    • 9 months ago
  20. kryton1212
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    i am still confusing..

    • 9 months ago
  21. Jhannybean
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    No i wasn't referring to that.

    • 9 months ago
  22. hartnn
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    @kryton1212 where are you confused ?

    • 9 months ago
  23. hartnn
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    you got this much ? \(\large |\dfrac{x+1}{8}|<1\) ?

    • 9 months ago
  24. kryton1212
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    everywhere... i still cannot get what you mean..

    • 9 months ago
  25. kryton1212
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    what does | this line mean

    • 9 months ago
  26. hartnn
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    the absolute value sign

    • 9 months ago
  27. kryton1212
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    oh i see..

    • 9 months ago
  28. hartnn
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    \(if \quad |a|<b \\ a<b \quad or \quad a>-b\)

    • 9 months ago
  29. kryton1212
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    okay..

    • 9 months ago
  30. hartnn
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    so, what about |(x+1)/8| <1 ?

    • 9 months ago
  31. kryton1212
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    what does it mean

    • 9 months ago
  32. Jhannybean
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    OH its an OR. OK that makes more sense now, Got it.

    • 9 months ago
  33. hartnn
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    \(if \quad |a|<b \\ a<b \quad or \quad a>-b\) \(\large if \quad |\dfrac{x+1}{8}|<1 \\ \large \dfrac{x+1}{8}<1 \quad or \quad \dfrac{x+1}{8}>-1\)

    • 9 months ago
  34. hartnn
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    now can you get the range ?

    • 9 months ago
  35. Jhannybean
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    So just solve for two x-values,and see which x-value fits the description of r<1.

    • 9 months ago
  36. kryton1212
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    -9<x<7

    • 9 months ago
  37. kryton1212
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    is it..

    • 9 months ago
  38. hartnn
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    yes, correct, thats your 1st part :)

    • 9 months ago
  39. kryton1212
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    thanks:) i got it ;D and part b.

    • 9 months ago
  40. hartnn
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    sum to infinity = \(\large \dfrac{a_1}{1-r}\)

    • 9 months ago
  41. hartnn
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    we have both, a1 and r

    • 9 months ago
  42. kryton1212
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    a1=?

    • 9 months ago
  43. hartnn
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    |dw:1372924494692:dw|

    • 9 months ago
  44. hartnn
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    \(\large a_1 =\dfrac{6(x+1)}{8}\)

    • 9 months ago
  45. hartnn
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    could you find the sum to infinity ?

    • 9 months ago
  46. kryton1212
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    |dw:1372925264670:dw|

    • 9 months ago
  47. hartnn
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    The general term in an geometric series is \(T(n)=a_1r^{n-1}\) where, a1 is the 1st term of the series (which we get when we put n=1 in the general term) and 'r' is the common ratio, (ratio of next term to current term)

    • 9 months ago
  48. kryton1212
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    a1 and r...okay

    • 9 months ago
  49. hartnn
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    if you want more details about geometric sequence, you can go through this when u have time, http://openstudy.com/study#/updates/503bb2a0e4b007f9003103b0

    • 9 months ago
  50. hartnn
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    any more doubts ? what u got as sum to infinity ?

    • 9 months ago
  51. kryton1212
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    |dw:1372925634484:dw|

    • 9 months ago
  52. Jhannybean
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    Yay!! now simplify :)

    • 9 months ago
  53. hartnn
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    |dw:1372925676770:dw|

    • 9 months ago
  54. kryton1212
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    oh yeah

    • 9 months ago
  55. kryton1212
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    6(x+1) / 7-x

    • 9 months ago
  56. hartnn
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    correct! now for the last part, the sum is given as 18 so, getting x from 6 (x+1) / (7-x) = 18 is just simple algebra :)

    • 9 months ago
  57. Jhannybean
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    \[\large 6 (\dfrac{x+1}{8})^n = 6 \dfrac{x+1}{8}(\dfrac{x+1}{8})^{n-1}\]\[\large a_{1}=\cfrac{6(x+1)}{8} \ \text{and} \ r = \left(\frac{x+1}{8}\right)\]\[\large s_n= \cfrac{ \cfrac{6(x+1)}{8}}{\cfrac{8-(x+1)}{8}}\]

    • 9 months ago
  58. hartnn
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    to give you a head start, multiply both sides by 7-x

    • 9 months ago
  59. Jhannybean
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    \[\large s_{n} = \cfrac{3(x+1)}{4}\cdot \cfrac{8}{-x+7} = \frac{6(x+1)}{-x+7}\]

    • 9 months ago
  60. kryton1212
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    x=5

    • 9 months ago
  61. kryton1212
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    thanks @hartnn and @Jhannybean :)

    • 9 months ago
  62. Jhannybean
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    So yeah...you equal that equation to 18 and solve for x :) You stated that x=5, let's check that! \[\large 18 = \cfrac{6(5+1)}{-5+7}\]\[\large 18=18 \ \checkmark \] good job!

    • 9 months ago
  63. hartnn
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    6x+6 = 126-18x 24x = 120 x=5 is correct :)

    • 9 months ago
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