SOLUTION: the first numbers p-1,2p-2 and 3p-1 are the first three terms of a Gp where p>0 find (i) the sum to infinity

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Question 1085372: the first numbers p-1,2p-2 and 3p-1 are the first three terms of a Gp where p>0 find
(i) the sum to infinity

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
GP means "geometric progression" which is another way to say "geometric sequence".

Term 1 = p-1
Term 2 = 2p-2
Term 3 = 3p-1

The common ratio r is the ratio of adajcent terms.
Specifically it's the result of dividing any term (but the first one) over its previous term.
The basic template is

common ratio = (any term but the first term)/(previous term)

Using that template, we can divide term 2 over term 1 to get

r = (term 2)/(term 1)
r = (2p-2)/(p-1)
r = (2(p-1))/(p-1)
r = 2

The (p-1) terms cancel out leaving r = 2, which is the common ratio.

Using the same template, we can divide term 3 over term 2 to get

r = (term 3)/(term 2)
r = (3p-1)/(2p-2)

Plug in r = 2 and solve for p.

r = (3p-1)/(2p-2)
2 = (3p-1)/(2p-2)
2(2p-2) = [(3p-1)/(2p-2)]*(2p-2)
2(2p-2) = 3p-1
4p-4 = 3p-1
4p-3p = -1+4
p = -1+4
p = 3

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We know the value of p is p = 3, so make substitutions to get

Term 1 = p-1 = 3-1 = 2
Term 2 = 2p-2 = 2*3-2 = 4
Term 3 = 3p-1 = 3*3-1 = 8

The first three terms are

Term 1 = 2
Term 2 = 4
Term 3 = 8

So the GP is 2,4,8,...

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This GP goes on forever and there is no limit to how big the terms grow.
This means that the sum to infinity is not a fixed number.
The sum will be infinity implying that it keeps growing forever (much like the terms do).
For this current problem, we cannot find the sum to infinity.

The only way to have the sum be a finite number, we need to have the common ratio r be between 0 and 1.
The value of r can't equal 0. The value of r can't equal 1.
Put another way we need 0+%3C+r+%3C+1 for the series to converge.
If r+%3E=+1, then the series diverges.