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kryton1212

  • one year 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
    • one year ago
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    \[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]

  2. kryton1212
    • one year ago
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    @hartnn @Callisto

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

  5. kryton1212
    • one year ago
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    umm..

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

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

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

  9. kryton1212
    • one year ago
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    ??

  10. hartnn
    • one year 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
    • one year ago
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    |dw:1372923030703:dw|

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

  13. kryton1212
    • one year ago
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    ahhh...

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

  15. primeralph
    • one year ago
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    @Jhannybean Why?

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

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

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

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

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

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

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

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

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

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

  26. hartnn
    • one year ago
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    the absolute value sign

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

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

  29. kryton1212
    • one year ago
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    okay..

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

  31. kryton1212
    • one year ago
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    what does it mean

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

  33. hartnn
    • one year 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
    • one year ago
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    now can you get the range ?

  35. Jhannybean
    • one year 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
    • one year ago
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    -9<x<7

  37. kryton1212
    • one year ago
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    is it..

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

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

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

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

  42. kryton1212
    • one year ago
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    a1=?

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

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

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

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

  47. hartnn
    • one year 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
    • one year ago
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    a1 and r...okay

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

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

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

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

  54. kryton1212
    • one year ago
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    oh yeah

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

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

  59. Jhannybean
    • one year 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
    • one year ago
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    x=5

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

  62. Jhannybean
    • one year 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
    • one year ago
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    6x+6 = 126-18x 24x = 120 x=5 is correct :)

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