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C2 Help!

Q9) i) The first, second and fourth terms of a geometric progression form the first 3 terms of an arithmetic progression. The terms are all different. Show that

r^3-2r+1=0

I got as far as:

a+ar+ar^3=a+(a+d)+(a+2d)

Then

How to I get rid of the ds?

Reply 1

vc94

If a, ar and ar^3 are the first 3 terms of an arithmetic sequence then they have the same common difference.

Can you form an equation with these terms?

Reply 2

sohail.s

Original post by vc94

If a, ar and ar^3 are the first 3 terms of an arithmetic sequence then they have the same common difference.

Can you form an equation with these terms?

thanks for that :smile:

, i didn't recognize that either :eek:

Reply 3

Plankey

I thought it meant the sum of the GP was equal to the sum of the AP terms. Instead, a=a; ar=a+d; ar^3=a+2d.

Reply 4

In One Ear

Original post by Plankey

I thought it meant the sum of the GP was equal to the sum of the AP terms. Instead, a=a; ar=a+d; ar^3=a+2d.

It also means that the sum of the GP is equal to the sum of the AP, its just that that is not helpful in solving the problem when examined like that.

Reply 5

hayzelle

Original post by Plankey

Q9) i) The first, second and fourth terms of a geometric progression form the first 3 terms of an arithmetic progression. The terms are all different. Show that

r^3-2r+1=0

I got as far as:

a+ar+ar^3=a+(a+d)+(a+2d)

Then

How to I get rid of the ds?

a+ar+ar^3=a+(a+d)+(a+2d)

a+ar+ar^3=3a+3d

a+d=ar
d=ar-a

sub in

a+ar+ar^3=3a+3(ar-a)
a+ar+ar^3=3a+3ar-3a
a+ar+ar^3=3ar
a+ar+ar^3=3ar
a-2ar+ar^3=0
rewrite neater: ar^3-2ar+a=0
divide by a: r^3-2r+1=0