the first term in a geometric series is 4 and the sum of the first three terms of the series is 64. find the sum of the first eight terms of the series.

This one is easier than the previous question you posted about GS's

Give this one a try, you can do it.

To find the sum of the first eight terms of a geometric series, we need to determine the common ratio (r) first.

Given that the first term is 4, we can label it as a₁. Additionally, we know that the sum of the first three terms of the series is 64. We can represent this as:

a₁ + a₁r + a₁r² = 64

Since the first term is 4, we can substitute a₁ with 4:

4 + 4r + 4r² = 64

Simplifying the equation, we get:

4r² + 4r - 60 = 0

Next, we can factor this quadratic equation:

(r - 3)(r + 5) = 0

This gives us two possible values for the common ratio:

r = 3 or r = -5

Since a common ratio cannot be negative in a geometric series, we can discard the negative value. Therefore, the common ratio is 3.

Now that we have the first term (a₁ = 4) and the common ratio (r = 3), we can find the sum of the first eight terms using the formula:

S₈ = a₁ * (1 - r⁸) / (1 - r)

Substituting the values:

S₈ = 4 * (1 - 3⁸) / (1 - 3)

Simplifying further:

S₈ = 4 * (1 - 6561) / (1 - 3)
S₈ = 4 * (-6560) / (-2)
S₈ = 13120

Therefore, the sum of the first eight terms of the series is 13120.