Question

30 POINTS!!!!!!

Answer 1

Answer with explanation:

→Infinite Geometric Series

$a_{1}=960\\\\r=\frac{1}{4}$

→The geometric series having common ratio r, and first term a,can be written as:$a, ar,ar^2,ar^3,ar^4,.......$

→So, the geometric Series can be Written as:

$960, 960 *\frac{1}{4},960*[\frac{1}{4}]^2,960*[\frac{1}{4}]^3,......\\\\ 960,240,60,15,.....$

→Sum of Infinite geometric Series

=960+240+60+15+.......

$={S_{\text{Infinity}}=\sum_{n=1}^{\infty }960*r^{n-1}=\frac{a}{1-r}$

$=\frac{960}{1-\frac{1}{4}}\\\\=\frac{960*4}{3}\\\\=320*4=1280\\\\=\frac{a}{1-r}$

Sum ,to infinity, Which is upper limit of this population=1280

Answer 2

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 = $\frac{1}{4}$

So, We have to write the sum in sigma notation:

$\sum (ar^{n-1})\\\\=\sum 960(\frac{1}{4})^{n-1}$

Since $r=\frac{1}{4}$

so, the sum is convergent, then,

$\sum 960(\frac{1}{4})^{n-1}=\frac{a}{1-r}=\frac{960}{1-\frac{1}{4}}=\frac{960}{\frac{3}{4}}=\frac{960\times 4}{3}=320\times 4=1280$

Hence, the sum that will be the upper limit of this population is 1280.