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?

2 Answers
Dean R.
May 8, 2018

S = a/{1-r} = {r^2-3r+2}/{1-r} = { (r-1)(r-2) }/{1-r} = 2-rS=a1−r=r2−3r+21−r=(r−1)(r−2)1−r=2−r

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

Explanation:

We have

S = sum_{k=0}^{infty} (r^2-3r+2) r^k S=∞∑k=0(r2−3r+2)rk

The general sum of an infinite geometric series is

sum_{k=0}^{infty} a r^k = a/{1-r}∞∑k=0ark=a1−r

In our case,

S = {r^2-3r+2}/{1-r} = { (r-1)(r-2) }/{1-r} = 2-rS=r2−3r+21−r=(r−1)(r−2)1−r=2−r

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

1 < S < 3 1<S<3

Somebody N.
May 8, 2018

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

Explanation:

ar^(n-1)arn−1

Where bbrr is the common ratio, bbaa is the first term and bbnn is the nth term.

We are told common ratio is rr

First term is (r^2-3r+2)(r2−3r+2)

The sum of a geometric series is given as:

a((1-r^n)/(1-r))a(1−rn1−r)

For the sum to infinity this simplifies to:

a/(1-r)a1−r

We are told this sum is S.

Substituting in our values for a and r:

(r^2-3r+2)/(1-r)=Sr2−3r+21−r=S

Factor the numerator:

((r-1)(r-2))/(1-r)=S(r−1)(r−2)1−r=S

Multiply numerator and denominator by -1−1

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

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