In a geometric sequence, consecutive terms have a common ratio:
![(a_(n+1))/(a_n)=r\quad \forall n\geq 1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/3997swpzbwqsuvqtiapohnl685trjit1fp.png)
This means that, in order to build a gometric sequence, we must choose an initial value
and a common ratio
, and we'll multiply each term by
to get the next one:
![a_1=a_1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/fltd5fa0b38x01bd966hj2eh99w0lo41m5.png)
![a_2=ra_1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/p9cd99g58n2gpft59ijcz1fzbtkuyj5kqp.png)
![a_3=r^2a_1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/4xlbmoy8pr4z1jz32zcuojt6dfe4z0xu8x.png)
![a_4=r^3a_1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/3p9wz3wuiajx8ddevnv1flwoy63chfzv04.png)
![a_5=r^4a_1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/2eev9qhmvmybbisk4vwmzwj71uq19pim31.png)
This implies that
![(a_5)/(a_2)=r^3](https://img.qammunity.org/2021/formulas/mathematics/middle-school/1hcjt4arcg9wa8rxlzkhhk8j5pb30tkybx.png)
And so in this case we have
![(10)/(80)=r^3 \iff r^3 = (1)/(8) \iff r=(1)/(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/wfswtcn5yocmibhikwgpr40h35oq3tbiln.png)