Answer: The sum that will be the upper limit of this population is 1280.
Step-by-step explanation:
Since we have given that
Initial population a₁ = 960
Common ratio = [tex]\frac{1}{4}[/tex]
So, We have to write the sum in sigma notation:
[tex]\sum (ar^{n-1})\\\\=\sum 960(\frac{1}{4})^{n-1}\\\\[/tex]
Since [tex]r=\frac{1}{4}<1[/tex]
so, the sum is convergent, then,
[tex]\sum 960(\frac{1}{4})^{n-1}=\frac{a}{1-r}=\frac{960}{1-\frac{1}{4}}=\frac{960}{\frac{3}{4}}=\frac{960\times 4}{3}=320\times 4=1280[/tex]
Hence, the sum that will be the upper limit of this population is 1280.
Answer with explanation:
→Infinite Geometric Series
[tex]a_{1}=960\\\\r=\frac{1}{4}[/tex]
→The geometric series having common ratio r, and first term a,can be written as:[tex]a, ar,ar^2,ar^3,ar^4,.......[/tex]
→So, the geometric Series can be Written as:
[tex]960, 960 *\frac{1}{4},960*[\frac{1}{4}]^2,960*[\frac{1}{4}]^3,......\\\\ 960,240,60,15,.....[/tex]
→Sum of Infinite geometric Series
=960+240+60+15+.......
[tex]={S_{\text{Infinity}}=\sum_{n=1}^{\infty }960*r^{n-1}=\frac{a}{1-r}[/tex]
[tex]=\frac{960}{1-\frac{1}{4}}\\\\=\frac{960*4}{3}\\\\=320*4=1280\\\\=\frac{a}{1-r}[/tex]
Sum ,to infinity, Which is upper limit of this population=1280
