Answer:
41.333
Step-by-step explanation:
Given is a geometric sequence
[tex]\frac{4}{3} ,\frac{8}{3} ,\frac{16}{3} ,\frac{32}{3} ,\frac{64}{3}[/tex]
No of terms n =5
a = I term =[tex]\frac{4}{3}[/tex]
Use the formula for sum of geometric sequence when r >1
[tex]S_n =\frac{a(r^n-1}{r-1} \\=\frac{4}{3}[\frac{ (2^5-1}{ 2 -1}] \\=4(\frac{31}{3} )\\=41.3333[/tex]