Question

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

Answer 1

The correct answer is C) t₁ = 375, $t_n=5t_{n-1}$.

From the general form,
$t_n=75(5)^n$, we must work backward to find t₁.

The general form is derived from the explicit form, which is
$t_n=t_1(r)^{n-1}$.  We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

$t_n=t_1(5)^{n-1}$

Using the products of exponents, we can "split up" the exponent:
$t_n=t_1(5)^n(5)^{-1}$

We know that 5⁻¹ = 1/5, so this gives us
$t_n=t_1(\frac{1}{5})(5)^n \\ \\=\frac{t_1}{5}(5)^n$

Comparing this to our general form, we see that
$\frac{t_1}{5}=75$

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
$t_n=t_{n-1}(r)$, while we must state what t₁ is; this gives us

$t_1=375; t_n=t_{n-1}(5)$