Find a general form of an equation of the line through the point a that satisfies the given condition. a(6, −5); parallel to the
1 answer:
You might be interested in
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
From the general form,
The general form is derived from the explicit form, which is
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