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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.
 9 months ago
 9 months 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.
 9 months ago
 9 months ago

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kryton1212Best ResponseYou've already chosen the best response.0
\[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
the sum to infinity will converge(exist) if \(r<1\) can you find 'r' from your general term ?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
general equation, \(T(n)=a_1r^{n1}\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
find a1 and 'r' both from your general term, you will need both
 9 months ago

primeralphBest ResponseYou've already chosen the best response.0
dw:1372921816382:dw
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
compare, try to make the power as (n1) \(\large 6 (\dfrac{x+1}{8})^n = 6 \dfrac{x+1}{8}(\dfrac{x+1}{8})^{n1}\) now compare this with a1 r^{n1} find a1 and r
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
then for the range of values of x, just do r <1
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
ohh..because @primeralph wrote "and" there! its actually "or"
 9 months ago

primeralphBest ResponseYou've already chosen the best response.0
dw:1372923436567:dw
 9 months ago

primeralphBest ResponseYou've already chosen the best response.0
@Jhannybean The x is only a part of r.
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Oh nevermind.... I was looking at it the wrong way.
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.0
i am still confusing..
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
No i wasn't referring to that.
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
@kryton1212 where are you confused ?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
you got this much ? \(\large \dfrac{x+1}{8}<1\) ?
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.0
everywhere... i still cannot get what you mean..
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.0
what does  this line mean
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
the absolute value sign
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
\(if \quad a<b \\ a<b \quad or \quad a>b\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
so, what about (x+1)/8 <1 ?
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
OH its an OR. OK that makes more sense now, Got it.
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
\(if \quad a<b \\ a<b \quad or \quad a>b\) \(\large if \quad \dfrac{x+1}{8}<1 \\ \large \dfrac{x+1}{8}<1 \quad or \quad \dfrac{x+1}{8}>1\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
now can you get the range ?
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
So just solve for two xvalues,and see which xvalue fits the description of r<1.
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
yes, correct, thats your 1st part :)
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.0
thanks:) i got it ;D and part b.
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
sum to infinity = \(\large \dfrac{a_1}{1r}\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
\(\large a_1 =\dfrac{6(x+1)}{8}\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
could you find the sum to infinity ?
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.0
dw:1372925264670:dw
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
The general term in an geometric series is \(T(n)=a_1r^{n1}\) 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)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
if you want more details about geometric sequence, you can go through this when u have time, http://openstudy.com/study#/updates/503bb2a0e4b007f9003103b0
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
any more doubts ? what u got as sum to infinity ?
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.0
dw:1372925634484:dw
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Yay!! now simplify :)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
correct! now for the last part, the sum is given as 18 so, getting x from 6 (x+1) / (7x) = 18 is just simple algebra :)
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
\[\large 6 (\dfrac{x+1}{8})^n = 6 \dfrac{x+1}{8}(\dfrac{x+1}{8})^{n1}\]\[\large a_{1}=\cfrac{6(x+1)}{8} \ \text{and} \ r = \left(\frac{x+1}{8}\right)\]\[\large s_n= \cfrac{ \cfrac{6(x+1)}{8}}{\cfrac{8(x+1)}{8}}\]
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
to give you a head start, multiply both sides by 7x
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
\[\large s_{n} = \cfrac{3(x+1)}{4}\cdot \cfrac{8}{x+7} = \frac{6(x+1)}{x+7}\]
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.0
thanks @hartnn and @Jhannybean :)
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
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!
 9 months ago

hartnnBest ResponseYou've already chosen the best response.3
6x+6 = 12618x 24x = 120 x=5 is correct :)
 9 months ago
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