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Help, OCR maths core 2 geometric progression

Hi, I need help answering all of the question below. It is from OCR maths core 2 may 2010.

The question is:

A geometric progression has first term a and common ratio r, and the terms are all different. The first, second and fourth terms of the geometric progression form the first three terms of an arithmetic progression.
(i) Show that r^3 −2r+1=0. [3]
(ii) Given that the geometric progression converges, find the exact value of r. [5]
(iii) Given also that the sum to infinity of this geometric progression is 3 + (square root) 5, find t (4)

Any help would be appreciated

Thanks
Reply 1
Avatar for raheem94
raheem94
13
Original post by cricket001
Hi, I need help answering all of the question below. It is from OCR maths core 2 may 2010.

The question is:

A geometric progression has first term a and common ratio r, and the terms are all different. The first, second and fourth terms of the geometric progression form the first three terms of an arithmetic progression.
(i) Show that r^3 −2r+1=0. [3]
(ii) Given that the geometric progression converges, find the exact value of r. [5]
(iii) Given also that the sum to infinity of this geometric progression is 3 + (square root) 5, find t (4)

Any help would be appreciated

Thanks


Part a,
The first 4 terms are, a,ar,ar2,ar3 a, ar, ar^2, ar^3

The terms that form a arithmetic progression are, a,ar,ar3 a, ar, ar^3

We know the difference between terms in arithmetic progression is constant, so the equation is, ar−a=ar3−ar ar-a = ar^3 -ar
Reply 2
Avatar for raheem94
raheem94
13
Original post by cricket001
Hi, I need help answering all of the question below. It is from OCR maths core 2 may 2010.

The question is:

A geometric progression has first term a and common ratio r, and the terms are all different. The first, second and fourth terms of the geometric progression form the first three terms of an arithmetic progression.
(i) Show that r^3 −2r+1=0. [3]
(ii) Given that the geometric progression converges, find the exact value of r. [5]
(iii) Given also that the sum to infinity of this geometric progression is 3 + (square root) 5, find t (4)

Any help would be appreciated

Thanks


Part b,

Geometric progression converges when |r|<1.
f(x)=r3−2r+1 f(x) = r^3 - 2r +1

Find one factor of f(x), by substituting in some values. Then use that to find all the factors, the required value of 'r' is the one which satisfies ∣r∣<1 |r|<1
But how do you get that into the form of r^3-2r 1=0
Original post by Stellafisher17
But how do you get that into the form of r^3-2r 1=0


From ar−a=ar3−arar-a = ar^3 -ar just divide through aa and rearrange into it.
Genius!! Thank you!

(Original post by RDKGames)From ar−a=ar3−arar-a = ar^3 -ar just divide through aa and rearrange into it.

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