Answer:
32.59 (nearest hundredth)
Step-by-step explanation:
Geometric sequence
General form of a geometric sequence: [tex]a_n=ar^{n-1}[/tex]
(where a is the first term and r is the common ratio)
Given:
[tex]\displaystyle \sum^{20}_{n=1} 4 \left(\dfrac{8}{9}\right)^{n-1}[/tex]
Therefore:
Sum of the first n terms of a geometric series:
[tex]S_n=\dfrac{a(1-r^n)}{1-r}[/tex]
To find the sum of the first 20 terms, substitute the found values of a and r, together with n = 20, into the formula:
[tex]\implies S_{20}=\dfrac{4\left(1-\left(\frac{8}{9}\right)^{20}\right)}{1-\left(\frac{8}{9}\right)}[/tex]
[tex]\implies S_{20}=32.58609013...[/tex]
[tex]\implies S_{20}=32.59\:\: \sf (nearest\:hundredth)[/tex]
General form of geometric progression
On comparing to the summation
Apply Sum formula
[tex]\boxed{\sf S_n=\dfrac{a(1-r^n)}{1-r}}[/tex]
[tex]\\ \implies \sf S_{20}=\dfrac{4\left(1-\left(\frac{8}{9}\right)^{20}\right)}{1-\left(\frac{8}{9}\right)}[/tex]
[tex]\\ \implies\sf S_{20}=32.586[/tex]
[tex]\\ \sf\mplies S_{20}=32.59[/tex]
