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kryton1212

  • 2 years ago

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.

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

  2. kryton1212
    • 2 years ago
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    @hartnn @Callisto

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

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

  5. kryton1212
    • 2 years ago
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    umm..

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

  7. kryton1212
    • 2 years ago
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    'r' is the ratio?

  8. primeralph
    • 2 years ago
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    |dw:1372921816382:dw|

  9. kryton1212
    • 2 years ago
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    ??

  10. hartnn
    • 2 years ago
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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

  11. hartnn
    • 2 years ago
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    |dw:1372923030703:dw|

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

  13. kryton1212
    • 2 years ago
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    ahhh...

  14. hartnn
    • 2 years ago
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    @Jhannybean what! ?

  15. primeralph
    • 2 years ago
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    @Jhannybean Why?

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

  17. primeralph
    • 2 years ago
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    |dw:1372923436567:dw|

  18. primeralph
    • 2 years ago
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    @Jhannybean The x is only a part of r.

  19. Jhannybean
    • 2 years ago
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    Oh nevermind.... I was looking at it the wrong way.

  20. kryton1212
    • 2 years ago
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    i am still confusing..

  21. Jhannybean
    • 2 years ago
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    No i wasn't referring to that.

  22. hartnn
    • 2 years ago
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    @kryton1212 where are you confused ?

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

  24. kryton1212
    • 2 years ago
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    everywhere... i still cannot get what you mean..

  25. kryton1212
    • 2 years ago
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    what does | this line mean

  26. hartnn
    • 2 years ago
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    the absolute value sign

  27. kryton1212
    • 2 years ago
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    oh i see..

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

  29. kryton1212
    • 2 years ago
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    okay..

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

  31. kryton1212
    • 2 years ago
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    what does it mean

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

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

  34. hartnn
    • 2 years ago
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    now can you get the range ?

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

  36. kryton1212
    • 2 years ago
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    -9<x<7

  37. kryton1212
    • 2 years ago
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    is it..

  38. hartnn
    • 2 years ago
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    yes, correct, thats your 1st part :)

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

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

  41. hartnn
    • 2 years ago
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    we have both, a1 and r

  42. kryton1212
    • 2 years ago
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    a1=?

  43. hartnn
    • 2 years ago
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    |dw:1372924494692:dw|

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

  45. hartnn
    • 2 years ago
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    could you find the sum to infinity ?

  46. kryton1212
    • 2 years ago
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    |dw:1372925264670:dw|

  47. hartnn
    • 2 years ago
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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)

  48. kryton1212
    • 2 years ago
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    a1 and r...okay

  49. hartnn
    • 2 years ago
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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

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

  51. kryton1212
    • 2 years ago
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    |dw:1372925634484:dw|

  52. Jhannybean
    • 2 years ago
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    Yay!! now simplify :)

  53. hartnn
    • 2 years ago
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    |dw:1372925676770:dw|

  54. kryton1212
    • 2 years ago
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    oh yeah

  55. kryton1212
    • 2 years ago
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    6(x+1) / 7-x

  56. hartnn
    • 2 years ago
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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 :)

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

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

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

  60. kryton1212
    • 2 years ago
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    x=5

  61. kryton1212
    • 2 years ago
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    thanks @hartnn and @Jhannybean :)

  62. Jhannybean
    • 2 years ago
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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!

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

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