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Sequence problem

  • Thread starter Numbnut247
  • Start date
  • #1
26
0
Hey guys:cool: , I need help for sum of a geometric sequence problem:

The first and second terms of a geometric sequence have a sum of 15, while the second and third terms have a sum of 60. Use an algebraic method to find the three terms.

This is what I have so far:

a + b + c
a + b = 15
b + c = 60
a = a
r = b/a
S2 = 15 = a(1-(b/a)^2)/1-(b/a)
S3 = 60 = a(1-(b/a)^3)/1-(b/a)

I then solved for a and b and got a = 3.75 and b = 11.25.

After knowing a and b, I find the common ratio: 3. But my numbers do not work for S3 because I got 48.75:grumpy: (Which, interestly enough is 11.25 away from 60:uhh: )

I'm really confused:confused: :cry:

Thanks
 

Answers and Replies

  • #2
1,074
1
Ok I just figured it out, lets get rid of the a, b, and c stuff and just deal with a and r. well you have 3 terms that differ by a power of r you have
{ar0, ar1, ar2}

you know that

a + ar = 15

and that

ar + ar2 = 60

divide the second equation by 4 and you can then set the two equations equal to one another and you end up with a nice quadratic to find r.
 
  • #3
26
0
thanks man:smile:
it's actually so simple i never thought about that:blushing:
thanks again:smile:
 
  • #4
1,074
1
Yea it took me a bit to realize that too, and I felt so stupid because I almost immediately realized what the asnwers had to be but couldn't figure out how to derive them for a bit.
 

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