# Another geometric series questionWatch

Thread starter 7 years ago
#1
Stuck on this:

The first 3 terms of a geometric series are the 5th, 9th and 12th terms of an arithmetic series that has common diff 2.

a) Find the common ratio

What ive gathered is that the first 3 terms:

a = a+4d
ar= a+8d
ar^2 =a+11d

Not sure if this is correct? But if so:

a=a+8
ar=a+16
ar^2 = a+22

Not sure what to do after that.

please help thanks!
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7 years ago
#2
You've assumed that the arithmetic series has the same first term as the geometric sequence. If that were true then your first equation gives , so that can't be true.

Suppose instead that the geometric sequence has first term and the arithmetic sequence has first term . Then you have:

One way of doing this is to look at the difference between the terms. The difference between the first two terms is 8, and between the second two is 6, and so and . You can solve these two equations simultaneously to find . [Hint: factorise the LHS of both equations and divide the second equation by the first.]
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Thread starter 7 years ago
#3
(Original post by nuodai)
You've assumed that the arithmetic series has the same first term as the geometric sequence. If that were true then your first equation gives , so that can't be true.

Suppose instead that the geometric sequence has first term and the arithmetic sequence has first term . Then you have:

One way of doing this is to look at the difference between the terms. The difference between the first two terms is 8, and between the second two is 6, and so and . You can solve these two equations simultaneously to find . [Hint: factorise the LHS of both equations and divide the second equation by the first.]

I worked out the answer and got it right. But how not sure how you derived those two equations?

Thanks!
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Thread starter 7 years ago
#4
nvm, got it.

Thanks alot nuodai!
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