Respuesta :
Answer:
C
Step-by-step explanation:
The sum to infinity of a geometric series is
[tex]\frac{a_{1} }{1-r}[/tex] ( - 1 < r < 1 ) ← substitute values
S( ∞ ) = [tex]\frac{5}{1-\frac{1}{3} }[/tex] = [tex]\frac{5}{\frac{2}{3} }[/tex] = [tex]\frac{15}{2}[/tex]
The sum of the infinite geometric series is [tex]\frac{15}{2}[/tex]
What is infinite geometric series ?
The sum of all the terms of a geometric sequence is called infinite geometric series.
If a, ar, a[tex]r^{2}[/tex], ...., a[tex]r^{n-1}[/tex], .... is a geometric series, then a+ar+a[tex]r^{2}[/tex]+ ....+a[tex]r^{n-1}[/tex]+.... is the corresponding infinite geometric series.
where a is the 1st term & r is the common ratio between consecutive terms of the series.
Sum of infinite geometric series = [tex]\frac{a}{1-r}[/tex] ; |r|<1
What is the required sum of series ?
Given, 1st term ([tex]a_{1}[/tex]) of the infinite geometric series = 5
Common ratio (r) = 1/3
∴ Sum of infinite geometric series = [tex]\frac{a_{1} }{1-r}[/tex]
= [tex]\frac{5}{1-\frac{1}{3} }[/tex]
= [tex]\frac{5}{\frac{2}{3} }[/tex]
= [tex]\frac{15}{2}[/tex]
Learn more about geometric series here :
https://brainly.com/question/24643676
#SPJ2
