Solve a Geometric Sequence Problem | Sum of 15 & 60 | Algebraic Method

[Numbnut247]

Hey guys , 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 (Which, interestly enough is 11.25 away from 60 )

I'm really confused

Thanks

 

[d_leet]

Ok I just figured it out, let's 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.

 

[Numbnut247]

thanks man
it's actually so simple i never thought about that
thanks again

 

[d_leet]

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.