Question

What is the sum of the first five terms of a geometric series with a1 = 6 and r = 2?

Answer 1

The geometric sequence is given by:

`\begin{gathered} a_n=ar^(n-1) \\ where\colon \\ a=first_{\text{ }}term_{\text{ }}of_{\text{ }}the_{\text{ }}sequence \\ r=common_{\text{ }}ratio \end{gathered}`

so:

`\begin{gathered} a_1=6 \\ r=2 \\ a_1=a\cdot2^(1-1)=a\cdot2^0=a=6 \\ so\colon \\ a=6 \end{gathered}`

Therefore, the geometric sequence is:

`\begin{gathered} a_n=6\cdot2^(n-1) \\ \end{gathered}`

the sum of the first five terms of a geometric series will be:

`\sum ^5_(n\mathop=1)6\cdot2^(n-1)=6+12+24+48+96=186`

For n = 5

`a_5=6\cdot2^(5-1)=6\cdot2^4=6\cdot16=96`