Question
The following problem has been puzzling me for a couple of days (nb: this is not homework).
There exists two geometric sequences that sum to 9. The value of their second term (t2) is 2.
- Find the common ratio (r)
- Find the first element (t1) of each
The answers to (1) are 2/3 and 1/3 and the answers to (2) are 3 and 6 respectively. Unfortunately, I can't understand how these were derived.
In tackling (1) I've tried to apply algebraic substitution to solve for r as follows:
t2 = t1*r; since t2 = 2 we have:
t1 = 2/r
The equation for calculating the sum (S) of a sequence that converges to a limit is given by:
S = t1 / (1 - r)
So, I tried to plug my value of t1 into S and solve for r as follows:
9 = (2/r) / (1-r)
9(1-r) = 2/r
2/9 = r(1-r)
Unfortunately, from this point I get stuck. I need to eliminate one of the r's but I can't seem to be able to.
Next, I thought to solve for r using the formula that sums the first 2 terms (S2) of the sequence:
S2 = (t1 (1-r^2)) / (1-r)
t1 + 2 = (t1 (1-r^2)) / (1-r)
but expanding this out I again run into the same problem (can't eliminate one of the r's).
So I have 2 questions:
- What am I doing wrong when deriving r?
- Once I have one of its values, how I derive the other?
Solution
2/9 = r(1-r)
Unfortunately, from this point I get stuck. I need to eliminate one of the r's but I can't seem to be able to.
You need to learn to factorise!
2/9 = r(1-r)
2/9 = r - r^2
2 = 9r - 9r^2
9r^2 - 9r + 2 = 0
(3r)^2 - 3(3r) + 2 = 0
to make it easier, let R = 3r
R^2 - 3R + 2 = 0
(R - 1)(R - 2) = 0
so 3r - 1 = 0, or 3r - 2 = 0
i.e. r = 1/3 or r = 2/3.
And your first term is 2/(1/3) = 6, or 2/(2/3) = 3
QED!
OTHER TIPS
2/9 = r (1 - r)
Rewrite this as ax2 + bx + c and use the quadratic formula to solve it:
2/9 = r - r2
r2 - r + 2/9 = 0
Using the quadratic formula, the roots are:
[-1 ± √(1 - 8/9)] / 2
= (1 ± 1/3) / 2
= 1/2 ± 1/6
= 1/3 or 2/3
Edit: Aw shoot, I spent way too long figuring out how to write plus/minus and square root. :-P