# Tricky One Here!

Printable View

• Jan 14th 2008, 03:58 PM
Gasser666
Tricky One Here!
Hey guys, (and girls), I'm new here and I was wondering if someone could get me started on this problem:

Find y so that (4y+1)+(y+4)+(10-y) form a geometric series.

So far, I know the following:

T1=4y+1
T2=y+4
T3=10-y

Tn=ar^(n-1)

Thx in advance!

EDIT: I also know that I need to find 2 anwsers for r.
• Jan 14th 2008, 04:08 PM
mr fantastic
Quote:

Originally Posted by Gasser666
Hey guys, (and girls), I'm new here and I was wondering if someone could get me started on this problem:

Find y so that (4y+1)+(y+4)+(10-y) form a geometric series.

So far, I know the following:

T1=4y+1
T2=y+4
T3=10-y

Tn=ar^(n-1)

Thx in advance!

EDIT: I also know that I need to find 2 anwsers for r.

You would know that for a geometric series $r = \frac{T_{n+1}}{T_n}$. In other words, the ratio of a term to the previous term is r.

So you require $r = \frac{y+4}{4y+1} = \frac{10-y}{y+4}$.

$\frac{y+4}{4y+1} = \frac{10-y}{y+4} \Rightarrow (y + 4)^2 = (4y + 1)(10 - y)$.

Expand, simplify and r-arrange into quadratic = 0. Solve the quadratic for y. Sub the value of y into either $r = \frac{y+4}{4y+1}$ or $r = \frac{10-y}{y+4}$ to get r.
• Jan 14th 2008, 04:14 PM
Gasser666
Thanks
Thanks alot, that seems alot easier now :)