Answer:
7.50 (nearest hundredth)
Step-by-step explanation:
General form of geometric progression: [tex]a_n=ar^{n-1}[/tex]
(where [tex]a[/tex] is the initial term and [tex]r[/tex] is the common ratio)
Given progression:
[tex]5\left(\dfrac13\right)^{n-1}[/tex]
Therefore:
- [tex]a=5[/tex]
- [tex]r=\dfrac13[/tex]
Sum of a geometric series:
[tex]S_n=\dfrac{a(1-r^n)}{(1-r)}[/tex]
Substituting a = 5, r = 1/3 and n = 10 to find the sum to n = 10:
[tex]\implies S_{10}=\dfrac{5(1-\frac13^{10})}{(1-\frac13)}[/tex]
[tex]\implies S_{10}=7.499873987...[/tex]
[tex]\implies S_{10}=7.50 \textsf{ (nearest hundredth)}[/tex]
