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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]
the sum to infinity will converge(exist) if \(|r|<1\) can you find 'r' from your general term ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question