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kryton1212
Group Title
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.
 one year ago
 one year ago
kryton1212 Group Title
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.
 one year ago
 one year ago

This Question is Closed

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
\[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
@hartnn @Callisto
 one year ago

hartnn Group TitleBest 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 ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
general equation, \(T(n)=a_1r^{n1}\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
find a1 and 'r' both from your general term, you will need both
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
'r' is the ratio?
 one year ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
dw:1372921816382:dw
 one year ago

hartnn Group TitleBest 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
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
dw:1372923030703:dw
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
then for the range of values of x, just do r <1
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
ahhh...
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
@Jhannybean what! ?
 one year ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
@Jhannybean Why?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
ohh..because @primeralph wrote "and" there! its actually "or"
 one year ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
dw:1372923436567:dw
 one year ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
@Jhannybean The x is only a part of r.
 one year ago

Jhannybean Group TitleBest ResponseYou've already chosen the best response.1
Oh nevermind.... I was looking at it the wrong way.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i am still confusing..
 one year ago

Jhannybean Group TitleBest ResponseYou've already chosen the best response.1
No i wasn't referring to that.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
@kryton1212 where are you confused ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
you got this much ? \(\large \dfrac{x+1}{8}<1\) ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
everywhere... i still cannot get what you mean..
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
what does  this line mean
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
the absolute value sign
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
oh i see..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
\(if \quad a<b \\ a<b \quad or \quad a>b\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
so, what about (x+1)/8 <1 ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
what does it mean
 one year ago

Jhannybean Group TitleBest ResponseYou've already chosen the best response.1
OH its an OR. OK that makes more sense now, Got it.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
\(if \quad a<b \\ a<b \quad or \quad a>b\) \(\large if \quad \dfrac{x+1}{8}<1 \\ \large \dfrac{x+1}{8}<1 \quad or \quad \dfrac{x+1}{8}>1\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
now can you get the range ?
 one year ago

Jhannybean Group TitleBest ResponseYou've already chosen the best response.1
So just solve for two xvalues,and see which xvalue fits the description of r<1.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
is it..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
yes, correct, thats your 1st part :)
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
thanks:) i got it ;D and part b.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
sum to infinity = \(\large \dfrac{a_1}{1r}\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
we have both, a1 and r
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
dw:1372924494692:dw
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
\(\large a_1 =\dfrac{6(x+1)}{8}\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
could you find the sum to infinity ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
dw:1372925264670:dw
 one year ago

hartnn Group TitleBest 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)
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
a1 and r...okay
 one year ago

hartnn Group TitleBest 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
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
any more doubts ? what u got as sum to infinity ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
dw:1372925634484:dw
 one year ago

Jhannybean Group TitleBest ResponseYou've already chosen the best response.1
Yay!! now simplify :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
dw:1372925676770:dw
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
oh yeah
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
6(x+1) / 7x
 one year ago

hartnn Group TitleBest 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 :)
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.3
to give you a head start, multiply both sides by 7x
 one year ago

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

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
thanks @hartnn and @Jhannybean :)
 one year ago

Jhannybean Group TitleBest 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!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
6x+6 = 12618x 24x = 120 x=5 is correct :)
 one year ago
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