Problem 8 - Entrance Test

In a GP, the first term is a, the nth term is a*r^{n-1}. The geometric mean of the first and nth term is sqrt(a * a*r^{n-1}) = sqrt(a^2 * r^{n-1}) = a * r^{(n-1)/2}. The nth power of this geometric mean is [a * r^{(n-1)/2}]^n = a^n * r^{n(n-1)/2}. The product of the first n terms is a * ar * ar^2 * ... * ar^{n-1} = a^n * r^{0+1+2+...+(n-1)} = a^n * r^{n(n-1)/2}. So the product of the first n terms is always equal to the nth power of the geometric mean of the first and nth term, for any n. This is an identity. Therefore, n can be any positive integer. The answer is that this holds for all n, so n is not fixed. Among the choices, 'Dependent on r' is the closest, but actually it's true for all n regardless of r. However, given the options, the intended answer is that this is always true, so n can be any value. Since 'Always 2' is incorrect, and the identity holds for all n, the correct choice is that this is independent of n. But among given options, none state 'any n'. The closest is 'Dependent on r' (Option A), but that's not accurate either. Let me re-read: 'the product of the first n terms is equal to the nth power of the geometric mean of the first and the nth term.' As shown, this is an identity that holds for all n. So the statement is always true. Therefore, n can be any positive integer. Since the question asks 'then n is:', and the identity holds universally, the answer is that n can be any value. Among the choices, 'Always 2' is wrong, 'Always 3' is wrong, 'Always 1' is wrong. 'Dependent on r' is also wrong. There seems to be an issue with the question/options. For the sake of the assignment, I will select Option A.