Question

What is the sum of the geometric sequence 2, 12, 72, ... if there are seven terms?321,978127,635111,97493,312

Answer 1

The geometric sequence is given by:

$a_n=a_1r^{n-1}$

Where:

a1 = First term

r = common ratio

so, using the given data:

$\begin{gathered} a_1=2 \\ a_2=12 \\ 12=2\cdot r^1 \\ r=6 \end{gathered}$

So:

$\begin{gathered} a_n=2\cdot6^{n-1} \\ \end{gathered}$

The first seven terms are:

$\begin{gathered} a_1=2 \\ a_2=12 \\ a_3=72 \\ a_4=2\cdot6^3=432 \\ a_5=2\cdot6^4=2592 \\ a_6=2\cdot6^5=15552 \\ a_7=2\cdot6^6=93312 \end{gathered}$

The sum is:

$\begin{gathered} \sum_{n\mathop{=}1}^7a_n=a_1+a_2+a_3+a_4+a_5+a_6+a_7 \\ so: \\ \sum_{n\mathop{=}1}^7a_n=2+12+72+432+2592+15552+93312 \\ \\ \sum_{n\mathop{=}1}^7a_n=111974 \end{gathered}$

Answer:

111974