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1. The common ratio of a geometric progression is r. The first term of the progression is r^2 − 3r + 2 and the sum to infinity is S.
Given S = 2 − r.
Find the set of possible values that S can take.
How do we find?
The common ratio of a geometric progression is r. The first term of the progression is r^2 − 3r + 2 and the sum to infinity is S.
Given S = 2 − r.
Find the set of possible values that S can take.
How do we find?
Given the common ratio, and the first term, then S=2-r, so being given that information is superfluous - it can't be anything else.

Can you link/scan/photo the actual question - what you've put doesn't look right.
The common ratio of a geometric progression is r. The first term of the progression is r^2 − 3r + 2 and the sum to infinity is S.
Given S = 2 − r.
Find the set of possible values that S can take.
How do we find?
We know that when the sum to infinity exists (i.e. if the sequence is convergent), it equals a/(1-r).
Thus (r^2 - 3r + 2)/(1-r) = 2-r, so
r^2 - 3r + 2 = (1-r)(2-r) = r^2 - 3r + 2,
So both sides are identical, and will always be equal as long as S is defined. However, know that the progression only converges and has a sum to infinity when -1 < r < 1.
Now we just have to find the range of S = 2-r, which will be 2-1<S<2-(-1) i.e. 1<S<3.
4. Oops my mistake of 'given'. Its says show that.
Q.2
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5. 9709_s17_qp_13.pdf (129.8 KB, 115 views)
Oops my mistake of 'given'. Its says show that.
Q.2
(Original post by ghostwalker)
Given the common ratio, and the first term, then S=2-r, so being given that information is superfluous - it can't be anything else.

Can you link/scan/photo the actual question - what you've put doesn't look right.
So, to show that S = 2-r, simply apply the formula that S=a/(1-r), i.e. S=(r^2-3r+2)/(1-r), but r^2-3r+2 = (r-1)(r-2) = (1-r)(2-r), since we can multiply both factors by -1 and this has the overall effect of multiplying by (-1)^2 = 1, leaving the answer unchanged. Hence we can now simply cancel the factor of (1-r) to give (2-r), as required.

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