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.

- anonymous

- chestercat

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- anonymous

\[T(n)=6(\frac{ x+1 }{ 8 })^{n}\]

- anonymous

@hartnn @Callisto

- 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

general equation, \(T(n)=a_1r^{n-1}\)

- anonymous

umm..

- hartnn

find a1 and 'r' both from your general term, you will need both

- anonymous

'r' is the ratio?

- primeralph

|dw:1372921816382:dw|

- anonymous

??

- 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

|dw:1372923030703:dw|

- hartnn

then for the range of values of x, just do |r| <1

- anonymous

ahhh...

- hartnn

@Jhannybean what! ?

- primeralph

@Jhannybean Why?

- hartnn

ohh..because @primeralph wrote "and" there!
its actually "or"

- primeralph

|dw:1372923436567:dw|

- primeralph

@Jhannybean The x is only a part of r.

- Jhannybean

Oh nevermind.... I was looking at it the wrong way.

- anonymous

i am still confusing..

- Jhannybean

No i wasn't referring to that.

- hartnn

@kryton1212 where are you confused ?

- hartnn

you got this much ?
\(\large |\dfrac{x+1}{8}|<1\)
?

- anonymous

everywhere...
i still cannot get what you mean..

- anonymous

what does | this line mean

- hartnn

the absolute value sign

- anonymous

oh i see..

- hartnn

\(if \quad |a|

**-b\)**

**
**

- anonymous

okay..

- hartnn

so, what about |(x+1)/8| <1 ?

- anonymous

what does it mean

- Jhannybean

OH its an OR. OK that makes more sense now, Got it.

- 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

now can you get the range ?

- Jhannybean

So just solve for two x-values,and see which x-value fits the description of r<1.

- anonymous

-9

- anonymous

is it..

- hartnn

yes, correct, thats your 1st part :)

- anonymous

thanks:) i got it ;D
and part b.

- hartnn

sum to infinity = \(\large \dfrac{a_1}{1-r}\)

- hartnn

we have both, a1 and r

- anonymous

a1=?

- hartnn

|dw:1372924494692:dw|

- hartnn

\(\large a_1 =\dfrac{6(x+1)}{8}\)

- hartnn

could you find the sum to infinity ?

- anonymous

|dw:1372925264670:dw|

- 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

a1 and r...okay

- 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

any more doubts ? what u got as sum to infinity ?

- anonymous

|dw:1372925634484:dw|

- Jhannybean

Yay!! now simplify :)

- hartnn

|dw:1372925676770:dw|

- anonymous

oh yeah

- anonymous

6(x+1) / 7-x

- 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

\[\large 6 (\dfrac{x+1}{8})^n = 6 \dfrac{x+1}{8}(\dfrac{x+1}{8})^{n-1}\]\[\large a_{1}=\cfrac{6(x+1)}{8} \ \text{and} \ r = \left(\frac{x+1}{8}\right)\]\[\large s_n= \cfrac{ \cfrac{6(x+1)}{8}}{\cfrac{8-(x+1)}{8}}\]

- hartnn

to give you a head start, multiply both sides by 7-x

- Jhannybean

\[\large s_{n} = \cfrac{3(x+1)}{4}\cdot \cfrac{8}{-x+7} = \frac{6(x+1)}{-x+7}\]

- anonymous

x=5

- anonymous

thanks @hartnn and @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

6x+6 = 126-18x
24x = 120
x=5 is correct :)

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