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kryton1212 Group Title

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.

  • one year ago
  • one year ago

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

    • one year ago
  2. kryton1212 Group Title
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    @hartnn @Callisto

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

    • one year ago
  4. hartnn Group Title
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    general equation, \(T(n)=a_1r^{n-1}\)

    • one year ago
  5. kryton1212 Group Title
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    umm..

    • one year ago
  6. hartnn Group Title
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    find a1 and 'r' both from your general term, you will need both

    • one year ago
  7. kryton1212 Group Title
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    'r' is the ratio?

    • one year ago
  8. primeralph Group Title
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    |dw:1372921816382:dw|

    • one year ago
  9. kryton1212 Group Title
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    ??

    • one year ago
  10. hartnn Group Title
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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

    • one year ago
  11. hartnn Group Title
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    |dw:1372923030703:dw|

    • one year ago
  12. hartnn Group Title
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    then for the range of values of x, just do |r| <1

    • one year ago
  13. kryton1212 Group Title
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    ahhh...

    • one year ago
  14. hartnn Group Title
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    @Jhannybean what! ?

    • one year ago
  15. primeralph Group Title
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    @Jhannybean Why?

    • one year ago
  16. hartnn Group Title
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    ohh..because @primeralph wrote "and" there! its actually "or"

    • one year ago
  17. primeralph Group Title
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    |dw:1372923436567:dw|

    • one year ago
  18. primeralph Group Title
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    @Jhannybean The x is only a part of r.

    • one year ago
  19. Jhannybean Group Title
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    Oh nevermind.... I was looking at it the wrong way.

    • one year ago
  20. kryton1212 Group Title
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    i am still confusing..

    • one year ago
  21. Jhannybean Group Title
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    No i wasn't referring to that.

    • one year ago
  22. hartnn Group Title
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    @kryton1212 where are you confused ?

    • one year ago
  23. hartnn Group Title
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    you got this much ? \(\large |\dfrac{x+1}{8}|<1\) ?

    • one year ago
  24. kryton1212 Group Title
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    everywhere... i still cannot get what you mean..

    • one year ago
  25. kryton1212 Group Title
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    what does | this line mean

    • one year ago
  26. hartnn Group Title
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    the absolute value sign

    • one year ago
  27. kryton1212 Group Title
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    oh i see..

    • one year ago
  28. hartnn Group Title
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    \(if \quad |a|<b \\ a<b \quad or \quad a>-b\)

    • one year ago
  29. kryton1212 Group Title
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    okay..

    • one year ago
  30. hartnn Group Title
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    so, what about |(x+1)/8| <1 ?

    • one year ago
  31. kryton1212 Group Title
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    what does it mean

    • one year ago
  32. Jhannybean Group Title
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    OH its an OR. OK that makes more sense now, Got it.

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

    • one year ago
  34. hartnn Group Title
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    now can you get the range ?

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

    • one year ago
  36. kryton1212 Group Title
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    -9<x<7

    • one year ago
  37. kryton1212 Group Title
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    is it..

    • one year ago
  38. hartnn Group Title
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    yes, correct, thats your 1st part :)

    • one year ago
  39. kryton1212 Group Title
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    thanks:) i got it ;D and part b.

    • one year ago
  40. hartnn Group Title
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    sum to infinity = \(\large \dfrac{a_1}{1-r}\)

    • one year ago
  41. hartnn Group Title
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    we have both, a1 and r

    • one year ago
  42. kryton1212 Group Title
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    a1=?

    • one year ago
  43. hartnn Group Title
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    |dw:1372924494692:dw|

    • one year ago
  44. hartnn Group Title
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    \(\large a_1 =\dfrac{6(x+1)}{8}\)

    • one year ago
  45. hartnn Group Title
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    could you find the sum to infinity ?

    • one year ago
  46. kryton1212 Group Title
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    |dw:1372925264670:dw|

    • one year ago
  47. hartnn Group Title
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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)

    • one year ago
  48. kryton1212 Group Title
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    a1 and r...okay

    • one year ago
  49. hartnn Group Title
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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

    • one year ago
  50. hartnn Group Title
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    any more doubts ? what u got as sum to infinity ?

    • one year ago
  51. kryton1212 Group Title
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    |dw:1372925634484:dw|

    • one year ago
  52. Jhannybean Group Title
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    Yay!! now simplify :)

    • one year ago
  53. hartnn Group Title
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    |dw:1372925676770:dw|

    • one year ago
  54. kryton1212 Group Title
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    oh yeah

    • one year ago
  55. kryton1212 Group Title
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    6(x+1) / 7-x

    • one year ago
  56. hartnn Group Title
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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 :)

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

    • one year ago
  58. hartnn Group Title
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    to give you a head start, multiply both sides by 7-x

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

    • one year ago
  60. kryton1212 Group Title
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    x=5

    • one year ago
  61. kryton1212 Group Title
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    thanks @hartnn and @Jhannybean :)

    • one year ago
  62. Jhannybean Group Title
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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!

    • one year ago
  63. hartnn Group Title
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    6x+6 = 126-18x 24x = 120 x=5 is correct :)

    • one year ago
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