Find the sum of the geometric sequence of the first eight term of 12 6 and 3

Answer 1

To find the sum of a geometric sequence, we can use the formula:

$$S_n = \frac{a(1 - r^n)}{1 - r}$$

where $S_n$ is the sum, $a$ is the first term, $n$ is the number of terms, and $r$ is the common ratio.

In this case, the first term $a = 12$, and we are given that the terms form a geometric sequence, so we can find the common ratio $r$ by dividing any term by the previous term:

$r = \frac{6}{12} = \frac{1}{2}$

We are asked for the sum of the first eight terms, so $n = 8$.

Plugging in these values into the formula, we have:

$$S_8 = \frac{12 \left(1 - \left(\frac{1}{2}\right)^8\right)}{1 - \frac{1}{2}}$$

Simplifying further:

$$S_8 = \frac{12 \left(1 - \frac{1}{256}\right)}{\frac{1}{2}}$$

$$S_8 = 12 \left(1 - \frac{1}{256}\right) \cdot 2$$

$$S_8 = 12 \cdot 2 - 12 \cdot \frac{1}{256} \cdot 2$$

$$S_8 = 24 - \frac{24}{128}$$

$$S_8 = 24 - \frac{3}{4}$$

Putting this into a common denominator and simplifying:

$$S_8 = \frac{96}{4} - \frac{3}{4}$$

$$S_8 = \frac{93}{4}$$

Therefore, the sum of the geometric sequence of the first eight terms is $\frac{93}{4}$.