Ok
We have our data:
![17;\text{ 187; 2057; 22627}](https://img.qammunity.org/2023/formulas/mathematics/college/3lon58rn3sdtqnjaw4a1jmgq5cfu8uz6fe.png)
We first find the ratio (r), and that we do dividing one of the values for the inmediatly lower value:
![r=(187)/(17)\Rightarrow r=11](https://img.qammunity.org/2023/formulas/mathematics/college/2itd2gfcil3a823l6jc48dubvjo7owxor7.png)
And we doublecheck it by doing the same with other values:
![r=(2057)/(187)\Rightarrow r=11](https://img.qammunity.org/2023/formulas/mathematics/college/fnbnyb6kxrdetqzwpx8e5ubombjhkmgp3u.png)
Now that we have the ratio (r), we the add the first term (a):
![a=17](https://img.qammunity.org/2023/formulas/mathematics/college/p6xq57h5do772t6cha56w9gfmj1wh105mg.png)
So, by definition the geometric sequence will be going as follows:
![a_n=a\cdot r^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/kdh1nfg44o7n3hhei5p88rtr7w69u81pyn.png)
Where a_n will be the geometric sequence, a the first term, r the ratio and n the ammount of terms, so:
![a_n=(17)(11)^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/k41sjhlatux8kny982zdrthto6smew2k60.png)
Now if you want to find the 3rd value in the sequence, you just replace n and so on:
![a_3=(17)(11)^(3-1)\Rightarrow a_3=2057](https://img.qammunity.org/2023/formulas/mathematics/college/15l80vyvof296kk37hzvcecaiera8u2rue.png)
And the fourth:
![a_4=(17)(11)^(4-1)\Rightarrow a_4=22627](https://img.qammunity.org/2023/formulas/mathematics/college/a7uo8bbu5yst9z08w1o9dwhco1ipi3dl4b.png)
Therefore, the geometric sequence is:
![a_n=(17)(11)^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/k41sjhlatux8kny982zdrthto6smew2k60.png)