The correct answer is C) t₁ = 375, [tex]t_n=5t_{n-1}[/tex].
From the general form,
[tex]t_n=75(5)^n[/tex], we must work backward to find t₁.
The general form is derived from the explicit form, which is
[tex]t_n=t_1(r)^{n-1}[/tex]. We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us
[tex]t_n=t_1(5)^{n-1}[/tex]
Using the products of exponents, we can "split up" the exponent:
[tex]t_n=t_1(5)^n(5)^{-1}[/tex]
We know that 5⁻¹ = 1/5, so this gives us
[tex]t_n=t_1(\frac{1}{5})(5)^n
\\
\\=\frac{t_1}{5}(5)^n[/tex]
Comparing this to our general form, we see that
[tex]\frac{t_1}{5}=75[/tex]
Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375
The recursive formula for a geometric sequence is given by
[tex]t_n=t_{n-1}(r)[/tex], while we must state what t₁ is; this gives us
[tex]t_1=375; t_n=t_{n-1}(5)[/tex]
