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.
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kryton1212
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\[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]
kryton1212
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@hartnn @Callisto
hartnn
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the sum to infinity will converge(exist) if \(|r|<1\)
can you find 'r' from your general term ?
hartnn
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general equation, \(T(n)=a_1r^{n-1}\)
kryton1212
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umm..
hartnn
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find a1 and 'r' both from your general term, you will need both
kryton1212
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'r' is the ratio?
primeralph
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|dw:1372921816382:dw|
kryton1212
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??
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
hartnn
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|dw:1372923030703:dw|
hartnn
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then for the range of values of x, just do |r| <1
kryton1212
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ahhh...
hartnn
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@Jhannybean what! ?
primeralph
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@Jhannybean Why?
hartnn
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ohh..because @primeralph wrote "and" there!
its actually "or"
primeralph
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|dw:1372923436567:dw|
primeralph
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@Jhannybean The x is only a part of r.
Jhannybean
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Oh nevermind.... I was looking at it the wrong way.
kryton1212
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i am still confusing..
Jhannybean
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No i wasn't referring to that.
hartnn
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@kryton1212 where are you confused ?
hartnn
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you got this much ?
\(\large |\dfrac{x+1}{8}|<1\)
?
kryton1212
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everywhere...
i still cannot get what you mean..
kryton1212
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what does | this line mean
hartnn
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the absolute value sign
kryton1212
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oh i see..
hartnn
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\(if \quad |a|<b \\ a<b \quad or \quad a>-b\)
kryton1212
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okay..
hartnn
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so, what about |(x+1)/8| <1 ?
kryton1212
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what does it mean
Jhannybean
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OH its an OR. OK that makes more sense now, Got it.
hartnn
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\(if \quad |a|<b \\ a<b \quad or \quad a>-b\)
\(\large if \quad |\dfrac{x+1}{8}|<1 \\ \large \dfrac{x+1}{8}<1 \quad or \quad \dfrac{x+1}{8}>-1\)
hartnn
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now can you get the range ?
Jhannybean
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So just solve for two x-values,and see which x-value fits the description of r<1.
kryton1212
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-9<x<7
kryton1212
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is it..
hartnn
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yes, correct, thats your 1st part :)
kryton1212
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thanks:) i got it ;D
and part b.
hartnn
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sum to infinity = \(\large \dfrac{a_1}{1-r}\)
hartnn
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we have both, a1 and r
kryton1212
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a1=?
hartnn
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|dw:1372924494692:dw|
hartnn
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\(\large a_1 =\dfrac{6(x+1)}{8}\)
hartnn
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could you find the sum to infinity ?
kryton1212
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|dw:1372925264670:dw|
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)
kryton1212
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a1 and r...okay
hartnn
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any more doubts ? what u got as sum to infinity ?
kryton1212
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|dw:1372925634484:dw|
Jhannybean
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Yay!! now simplify :)
hartnn
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|dw:1372925676770:dw|
kryton1212
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oh yeah
kryton1212
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6(x+1) / 7-x
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 :)
Jhannybean
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\[\large 6 (\dfrac{x+1}{8})^n = 6 \dfrac{x+1}{8}(\dfrac{x+1}{8})^{n-1}\]\[\large a_{1}=\cfrac{6(x+1)}{8} \ \text{and} \ r = \left(\frac{x+1}{8}\right)\]\[\large s_n= \cfrac{ \cfrac{6(x+1)}{8}}{\cfrac{8-(x+1)}{8}}\]
hartnn
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to give you a head start, multiply both sides by 7-x
Jhannybean
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\[\large s_{n} = \cfrac{3(x+1)}{4}\cdot \cfrac{8}{-x+7} = \frac{6(x+1)}{-x+7}\]
kryton1212
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x=5
kryton1212
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thanks @hartnn and @Jhannybean :)
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!
hartnn
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6x+6 = 126-18x
24x = 120
x=5 is correct :)