Final answer:
The conditions given lead to a contradiction, as a geometric progression with a common ratio (r) of 1 does not have a finite sum to infinity. Therefore, there is no solution for the first term of the GP under the given scenario.
Step-by-step explanation:
To find the first term of the geometric progression (GP) when the second, third, and first terms form an arithmetic progression (AP), and given that the sum to infinite terms of the GP is 36, let's denote the first term by a and the common ratio by r. Since the GP sum to infinity is given by S = a / (1 - r), and S = 36, we have:
a / (1 - r) = 36 ... (1)
For the second and third terms of the GP to form an AP with the first term:
And their order in the AP is a, ar, ar^2. The property of an AP is that twice the middle term is equal to the sum of the other two terms, so:
2ar = a + ar^2 ... (2)
Dividing by a (assuming a ≠ 0):
2r = 1 + r^2 ... (3)
This can be rearranged to the quadratic equation:
r^2 - 2r + 1 = 0 ... (4)
Solving for r gives r = 1. Substituting r = 1 in equation (1), we get:
a / (1 - 1) = 36, which cannot be true because division by zero is undefined. Therefore, there is no solution under the given conditions, as a sum to infinity for a GP with r = 1 would not be finite.