Question

The common ratio of a ggeometric progression is r the first term of the progression is(r^2-3r+2) and the sum of infinity is S Show that S=2-r (I have) Find the set of possible values that S can take?

Answer 1

`S = a/{1-r} = {r^2-3r+2}/{1-r} = { (r-1)(r-2) }/{1-r} = 2-r`

Since `|r|<1` we get `1 < S < 3`

Explanation:

We have

`S = sum_{k=0}^{infty} (r^2-3r+2) r^k`

The general sum of an infinite geometric series is

`sum_{k=0}^{infty} a r^k = a/{1-r}`

In our case,

`S = {r^2-3r+2}/{1-r} = { (r-1)(r-2) }/{1-r} = 2-r`

Geometric series only converge when `|r|<1`, so we get

`1 < S < 3`

Answer 2

`color(blue)(1 < S < 3)`

Explanation:

`ar^(n-1)`

Where `bbr` is the common ratio, `bba` is the first term and `bbn` is the nth term.

We are told common ratio is `r`

First term is `(r^2-3r+2)`

The sum of a geometric series is given as:

`a((1-r^n)/(1-r))`

For the sum to infinity this simplifies to:

`a/(1-r)`

We are told this sum is S.

Substituting in our values for a and r:

`(r^2-3r+2)/(1-r)=S`

Factor the numerator:

`((r-1)(r-2))/(1-r)=S`

Multiply numerator and denominator by `-1`

`((r-1)(2-r))/(r-1)=S`

Cancelling:

`(cancel((r-1))(2-r))/(cancel((1-r)))=S`

`S=2-r`

To find the possible values we remember that a geometric series only has a sum to infinity if `-1< r < 1`

`2-1 <2 -r <1+2`

`1 < 2-r < 3`

i.e.

`1 < S < 3`