In a geometric progression with terms, the sum of the first and last term is 66 and the product of the second and second last term is 128. Given that the sum of all the terms of this geometric progression is 126, find the number of terms and possible values of common ratio.
To solve this problem, let's begin by defining the terms in the geometric progression. Let the first term be "a" and the common ratio be "r".
Now, we can use the information provided to set up equations.
1. The sum of the first and last terms is 66:
a + ar^(n-1) = 66
2. The product of the second and second last terms is 128:
ar + ar^(n-2) = 128
3. The sum of all the terms is 126:
a(1 - r^n)/(1 - r) = 126
We have three equations and three unknowns (a, r, and n). We can solve these equations simultaneously to find the values.
Now, let's proceed with solving the equations.
From equation 1, we can rewrite it as:
a(1 + r^(n-1)) = 66
From equation 2, we can rewrite it as:
ar(1 + r^(n-3)) = 128
To eliminate a, we can divide equation 3 by equation 1:
(a(1 - r^n))/(a(1 + r^(n-1))) = 126/66
Simplifying equation 3, we have:
(1 - r^n)/(1 + r^(n-1)) = 2
Cross-multiplying, we get:
1 - r^n = 2(1 + r^(n-1))
Expanding, we have:
1 - r^n = 2 + 2r^(n-1)
Rearranging the terms, we get:
r^n + 2r^(n-1) - r^n = 1 - 2
Simplifying, we have:
2r^(n-1) = -1
Dividing by 2, we get:
r^(n-1) = -1/2
Since r^(n-1) is negative, we know that r must be negative.
Now, let's consider some values of r. Since r is negative, we will consider r = -1/2 and r = -2.
For r = -1/2, we have:
(-1/2)^(n-1) = -1/2
This equation has no real solutions.
For r = -2, we have:
(-2)^(n-1) = -1/2
Taking the logarithm of both sides, we have:
(n-1)log(-2) = log(-1/2)
Since log(-1/2) is not defined in the real number system, r = -2 does not yield a valid solution.
Therefore, there are no possible values for the common ratio and the number of terms that satisfy the given conditions.
So, suppose there are n terms. You know that
a + ar^(n-1) = 66
ar * ar^(n-2) = 128
a(r^n - 1)/(r-1) = 126
You can go through the math, but 128 is a product of powers of 2, so just looking at it, I'd guess the sequence is
2,4,8,16,32,64