suppose an arithmetic sequence and a geometric sequence with common ration r have the same first two terms. show that the third term of the geometric series is r^2/(2r-1) tomes the third term of the arithmetic sequence
the first 3 terms of the AS are a, a+d, a+2d
the first 3 terms of the GS are a, ar, ar^2
you said the second terms are equal
then a+d = ar
d = ar-a
so third term of GS/third term of AS
= ar^2/(a+2d)
= ar^2(a + 2(ar-a)
= ar^2/(a + 2ar - 2a)
= ar^2/(2ar - a) now divide top and bottom by a
= r^2/(2r-1)
swaggot swag
To prove that the third term of the geometric sequence is r^2 / (2r - 1) times the third term of the arithmetic sequence, we need to make use of the formulas for the nth term of an arithmetic sequence and the nth term of a geometric sequence.
Let's assume that the first term of both sequences is 'a' and the common ratio for the geometric sequence is 'r'. The arithmetic sequence can be represented as:
Arithmetic sequence: a, a + d, a + 2d, ...
where 'd' is the common difference between consecutive terms.
The geometric sequence can be represented as:
Geometric sequence: a, ar, ar^2, ...
Now, let's determine the first two terms of both sequences:
First term of the arithmetic sequence = a
Second term of the arithmetic sequence = a + d
First term of the geometric sequence = a
Second term of the geometric sequence = ar
Since the first two terms are the same for both sequences, we can set up the following equation:
a + d = ar
Now, let's calculate the third term of each sequence:
Third term of the arithmetic sequence = a + 2d
Third term of the geometric sequence = ar^2
To prove that the third term of the geometric sequence is r^2 / (2r - 1) times the third term of the arithmetic sequence, we want to show that:
ar^2 = (r^2 / (2r - 1)) * (a + 2d)
First, let's express 'd' in terms of 'a' and 'r' using the equation a + d = ar:
d = ar - a
Substituting this value into the expression for the third term of the arithmetic sequence, we can rewrite it as:
Third term of the arithmetic sequence = a + 2(ar - a)
= 3ar - 2a
Now, substituting the expressions for the third term of the arithmetic sequence and the third term of the geometric sequence into the equation we want to prove:
ar^2 = (r^2 / (2r - 1)) * (a + 2(ar - a))
Simplifying this expression:
ar^2 = (r^2 / (2r - 1)) * (3ar - 2a)
Next, multiply both sides of the equation by (2r - 1) to eliminate the fraction:
ar^2(2r - 1) = r^2(3ar - 2a)
Expanding both sides of the equation:
2ar^3 - ar^2 = 3ar^3 - 2r^2a
Rearranging the terms:
3ar^3 - 2ar^2 = 2r^2a
Factoring out 'a' from both sides:
a(3r^3 - 2r^2) = 2r^2a
Canceling out 'a' on both sides:
3r^3 - 2r^2 = 2r^2
Rearranging the equation:
3r^3 - 4r^2 = 0
Factoring out 'r^2':
r^2(3r - 4) = 0
Since we are assuming that 'r' is nonzero, we can divide by 'r^2':
3r - 4 = 0
Solving for 'r':
3r = 4
r = 4/3
Therefore, we have shown that when the common ratio 'r' is equal to 4/3, the third term of the geometric sequence is (4/3)^2 / (2(4/3) - 1) times the third term of the arithmetic sequence.
Please note that this proof is specific to the given equations and parameters.