A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequence? A) t1 = 75, tn = 5tn - 1, where n ∈N and n > 1 B) t1 = 75, tn = 75tn - 1, where n ∈N and n > 1 C) t1 = 375, tn = 5tn - 1, where n ∈N and n > 1 D) t1 = 375, tn = 5tn + 1, where n ∈N and n > 1
Accepted Solution
A:
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