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

the sum to infinity will converge(exist) if \(|r|<1\)
can you find 'r' from your general term ?

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|

??

|dw:1372923030703:dw|

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

ahhh...

@Jhannybean what! ?

@Jhannybean Why?

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..

okay..

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

what does it mean

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

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|

a1 and r...okay

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

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 :)

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