A geometric series has a positive common ratio of r. The series has a sum to infinity of 9 and the sum of the first two terms is 5. Find the first four terms of the series.
Again, use your definitions.
What two equations would you get ?
To find the first four terms of the geometric series, we need to determine the common ratio (r) and the first term (a).
We are given that the sum to infinity of the series is 9. In a geometric series, the sum to infinity is given by the formula: S = a / (1 - r), where S represents the sum to infinity, a is the first term, and r is the common ratio.
We are given that the sum of the first two terms is 5. Therefore, we can write the equation: a + ar = 5.
Now we have two equations:
1) a / (1 - r) = 9
2) a + ar = 5
We can solve these equations simultaneously to find the values of a and r.
From equation 2) we can rewrite it as: a(1 + r) = 5. Dividing both sides by (1 + r), we get: a = 5 / (1 + r).
Now we can substitute this expression for a into equation 1):
(5 / (1 + r)) / (1 - r) = 9
Simplifying this equation, we get: 5 / (1 - r + r^2) = 9
Cross-multiplying, we have: 9(1 - r + r^2) = 5
Expanding and rearranging, we get: 9r^2 - 9r + 4 = 0
This is a quadratic equation in terms of r. We can solve it using factoring, quadratic formula, or any other method of solving quadratic equations.
However, since we are only interested in the positive common ratio, we can determine the values of r by inspection. We notice that the factors of 4 (the constant term) that have a sum of -9 (the coefficient of the linear term) are -4 and -1. Therefore, the quadratic factors as (3r - 4)(3r - 1) = 0.
This gives us two possible values for r:
r1 = 4/3
r2 = 1/3
Now, for each possible value of r, we can substitute it back into equation 2) to find the corresponding value of a:
For r1 = 4/3:
a = 5 / (1 + 4/3) = 5 / (7/3) = 15 / 7
For r2 = 1/3:
a = 5 / (1 + 1/3) = 5 / (4/3) = 15 / 4
Therefore, the first term (a) and the common ratio (r) for the geometric series are:
For r1 = 4/3: a1 = 15/7, r1 = 4/3
For r2 = 1/3: a2 = 15/4, r2 = 1/3
To find the first four terms of the series, we can use the formula for the nth term of a geometric series: an = a * r^(n-1), where an is the nth term, a is the first term, r is the common ratio, and n is the term number.
For the first series (r1 = 4/3):
a1 = 15/7, r1 = 4/3
a1 = 15/7
a2 = (15/7) * (4/3) = 20/7
a3 = (15/7) * (4/3)^2 = 80/21
a4 = (15/7) * (4/3)^3 = 320/63
For the second series (r2 = 1/3):
a1 = 15/4, r2 = 1/3
a1 = 15/4
a2 = (15/4) * (1/3) = 5/4
a3 = (15/4) * (1/3)^2 = 5/36
a4 = (15/4) * (1/3)^3 = 5/108
Therefore, the first four terms of the geometric series for r1 = 4/3 are: 15/7, 20/7, 80/21, 320/63.
And the first four terms of the geometric series for r2 = 1/3 are: 15/4, 5/4, 5/36, 5/108.