Question

What is the sum of the geometric series in which a1 = 2, r = 3, and an = 486?

Answer 1

In order to determine the sum of the geometric series, proceed as follow:

Use the following expression for an:

$a_n=a_1r^{n-1}$

where,

a1 = 2

r = 3

an = 486

Replace the previous values into the expression for an, solve for n and simplify:

$\begin{gathered} 486=2\cdot3^{n-1} \\ \frac{486}{2}=3^{n-1} \\ 243=3^{n-1} \\ 3^5=3^{n-1} \end{gathered}$

Then, it is necessary that n = 6, because n - 1 = 6 - 1 = 5 and the previous equation is consistent.

Now, consider that the sum of the geometric series is given by:

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

Replace the values of the parameters and simplify:

$S_6=2(\frac{1-3^6}{1-3})=2(\frac{1-729}{-2})=728$

Hence, the result is 728