anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]
anonymous
  • anonymous
@hartnn @Callisto
hartnn
  • hartnn
the sum to infinity will converge(exist) if \(|r|<1\) can you find 'r' from your general term ?

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More answers

hartnn
  • hartnn
general equation, \(T(n)=a_1r^{n-1}\)
anonymous
  • anonymous
umm..
hartnn
  • hartnn
find a1 and 'r' both from your general term, you will need both
anonymous
  • anonymous
'r' is the ratio?
primeralph
  • primeralph
|dw:1372921816382:dw|
anonymous
  • anonymous
??
hartnn
  • hartnn
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
  • hartnn
|dw:1372923030703:dw|
hartnn
  • hartnn
then for the range of values of x, just do |r| <1
anonymous
  • anonymous
ahhh...
hartnn
  • hartnn
@Jhannybean what! ?
primeralph
  • primeralph
@Jhannybean Why?
hartnn
  • hartnn
ohh..because @primeralph wrote "and" there! its actually "or"
primeralph
  • primeralph
|dw:1372923436567:dw|
primeralph
  • primeralph
@Jhannybean The x is only a part of r.
Jhannybean
  • Jhannybean
Oh nevermind.... I was looking at it the wrong way.
anonymous
  • anonymous
i am still confusing..
Jhannybean
  • Jhannybean
No i wasn't referring to that.
hartnn
  • hartnn
@kryton1212 where are you confused ?
hartnn
  • hartnn
you got this much ? \(\large |\dfrac{x+1}{8}|<1\) ?
anonymous
  • anonymous
everywhere... i still cannot get what you mean..
anonymous
  • anonymous
what does | this line mean
hartnn
  • hartnn
the absolute value sign
anonymous
  • anonymous
oh i see..
hartnn
  • hartnn
\(if \quad |a|-b\)
anonymous
  • anonymous
okay..
hartnn
  • hartnn
so, what about |(x+1)/8| <1 ?
anonymous
  • anonymous
what does it mean
Jhannybean
  • Jhannybean
OH its an OR. OK that makes more sense now, Got it.
hartnn
  • hartnn
\(if \quad |a|-b\) \(\large if \quad |\dfrac{x+1}{8}|<1 \\ \large \dfrac{x+1}{8}<1 \quad or \quad \dfrac{x+1}{8}>-1\)
hartnn
  • hartnn
now can you get the range ?
Jhannybean
  • Jhannybean
So just solve for two x-values,and see which x-value fits the description of r<1.
anonymous
  • anonymous
-9
anonymous
  • anonymous
is it..
hartnn
  • hartnn
yes, correct, thats your 1st part :)
anonymous
  • anonymous
thanks:) i got it ;D and part b.
hartnn
  • hartnn
sum to infinity = \(\large \dfrac{a_1}{1-r}\)
hartnn
  • hartnn
we have both, a1 and r
anonymous
  • anonymous
a1=?
hartnn
  • hartnn
|dw:1372924494692:dw|
hartnn
  • hartnn
\(\large a_1 =\dfrac{6(x+1)}{8}\)
hartnn
  • hartnn
could you find the sum to infinity ?
anonymous
  • anonymous
|dw:1372925264670:dw|
hartnn
  • hartnn
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)
anonymous
  • anonymous
a1 and r...okay
hartnn
  • hartnn
if you want more details about geometric sequence, you can go through this when u have time, http://openstudy.com/study#/updates/503bb2a0e4b007f9003103b0
hartnn
  • hartnn
any more doubts ? what u got as sum to infinity ?
anonymous
  • anonymous
|dw:1372925634484:dw|
Jhannybean
  • Jhannybean
Yay!! now simplify :)
hartnn
  • hartnn
|dw:1372925676770:dw|
anonymous
  • anonymous
oh yeah
anonymous
  • anonymous
6(x+1) / 7-x
hartnn
  • hartnn
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
  • Jhannybean
\[\large 6 (\dfrac{x+1}{8})^n = 6 \dfrac{x+1}{8}(\dfrac{x+1}{8})^{n-1}\]\[\large a_{1}=\cfrac{6(x+1)}{8} \ \text{and} \ r = \left(\frac{x+1}{8}\right)\]\[\large s_n= \cfrac{ \cfrac{6(x+1)}{8}}{\cfrac{8-(x+1)}{8}}\]
hartnn
  • hartnn
to give you a head start, multiply both sides by 7-x
Jhannybean
  • Jhannybean
\[\large s_{n} = \cfrac{3(x+1)}{4}\cdot \cfrac{8}{-x+7} = \frac{6(x+1)}{-x+7}\]
anonymous
  • anonymous
x=5
anonymous
  • anonymous
thanks @hartnn and @Jhannybean :)
Jhannybean
  • Jhannybean
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
  • hartnn
6x+6 = 126-18x 24x = 120 x=5 is correct :)

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