Question

Find S3 of the sum of the geometric series. az = 4, a3 = 1, r= 1

Answer 1

Given:

a1 = 4

a3 = 1

r = ½

Let's find the S3 of the sum of the geometric series.

Apply the sum of geometric series formula below:

$S_n=\frac{a_1(1-r^n)}{1-r}$

Let's solve for S3.

Substitute the values into the equation.

Where: n = 3

Thus, we have:

$S_3=\frac{4(1-(\frac{1}{2})^3)^{}^{}}{1-\frac{1}{2}}$

Solving further:

$\begin{gathered} S_3=\frac{4(1-(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}))}{\frac{1}{2}} \\ \\ S_3=\frac{4(1-\frac{1}{8})}{\frac{1}{2}} \\ \\ S_3=\frac{4(\frac{7}{8})}{\frac{1}{2}} \\ \\ S_3=\frac{4\ast\frac{7}{8}}{\frac{1}{2}} \\ \\ S_3=\frac{\frac{7}{2}}{\frac{1}{2}} \\ \\ S_3=\frac{7}{2}\ast\frac{2}{1} \\ \\ S_3=7 \end{gathered}$

Therefore, the S3 of the sum of the given geometric series is 7

ANSWER:

7