The number of Fishes caught by the company over the first 8 years is 1,034,410 fishes
Here, we want to find the sum of the fishes the company was able to catch over the first 8 years
We start by setting up an exponential equation to represent the number of fishes caught at any number of years in time
We can have the exponential equation as;
![P(t)=I(1-r)^(t-1)](https://img.qammunity.org/2023/formulas/mathematics/college/6acnnfygegt424xqrvo4s01q27qj3rfcf5.png)
where I is the population of fishes in the first year
r is the percentage decrease which is 8% and it is same as 8/100 = 0.08
t is the year number
and P(t) is the population at a certain year
So substituting these values, we have;
![\begin{gathered} P(t)=170,000(1-0.08)^(t-1) \\ \\ P(t)=170,000(0.92)^(t-1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vmziew6d90wuty1x11yfvgd3bnzremhtt9.png)
So in this case, we have a geometric series with the nth term given above such that;
The first term a, is 170,000
The common ratio is 0.92
So the sum of the first 8 years which is the sum of the first 8 terms can be obtained using the formula for the geometric series as follows;
![\begin{gathered} S_n\text{ = }(a(1-r^n))/(1-r) \\ \\ S_8\text{ = }\frac{170,000(1-0.92^8)}{1-\text{ 0.92}} \\ \\ S_8\text{ =}(170,000(0.48678))/(0.08) \\ \\ S_8\text{ = }(82,752.792)/(0.08) \\ \\ S_8\text{ = 1,034,409.89} \\ \\ To\text{ the nearest integer, this is 1,034,410} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p0b963yufqcpl09lhr7fqg0nu8wzn9ef8n.png)