A geometric series is written as
, where
is the first term of the series and
is the common ratio.
In other words, to compute the next term in the series you have to multiply the previous one by
.
Since we know that the first time is 6 (but we don't know the common ratio), the first terms are
.
Let's use the other information, since the last term is
, we know that
, otherwise the terms would be bigger and bigger.
The information about the sum tells us that
![\displaystyle \sum_(i=0)^n 6r^i = 6\sum_(i=0)^n r^i = 6558](https://img.qammunity.org/2019/formulas/mathematics/middle-school/7vfqe3x4bgv4q3h0jafekp246hscj5arsx.png)
We have a formula to compute the sum of the powers of a certain variable, namely
![\displaystyle \sum_(i=0)^n r^i = \cfrac{r^(n+1)-1}{r-1}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/lv3pjf5bh7k7rb6xm93kerh204zaydblzb.png)
So, the equation becomes
![6\cfrac{r^(n+1)-1}{r-1} = 6558](https://img.qammunity.org/2019/formulas/mathematics/middle-school/f8hcfpsgi2vigp2mcctytm8616t8m8c041.png)
The only integer solution to this expression is
.
If you want to check the result, we have
![6+6*3+6*3^2+6*3^3+6*3^4+6*3^5+6*3^6 = 6558](https://img.qammunity.org/2019/formulas/mathematics/middle-school/zol114ylo0gy5trdd4k5phvl2urxvr53k9.png)
and the last term is
![6*3^6 = 4374](https://img.qammunity.org/2019/formulas/mathematics/middle-school/tn3hlmzu9qmcf6ven5p0hu6j4sd0oc4b77.png)