The correct answer is C) t₁ = 375,

.
From the general form,

, we must work backward to find t₁.
The general form is derived from the explicit form, which is

. We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

Using the products of exponents, we can "split up" the exponent:

We know that 5⁻¹ = 1/5, so this gives us

Comparing this to our general form, we see that

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375
The recursive formula for a geometric sequence is given by

, while we must state what t₁ is; this gives us