Question

2 A Series is given by 8+ 4 + 2 +

+--
27
9
3
a show that the series is a geometric progression
b Find the ninth term.

Answer 1

Hi,

Explanation:

`(4)/(9) =(8)/(27) *(3)/(2) \\\\(2)/(3) =(4)/(9) *(3)/(2) =(8)/(27)*{((3)/(2))}^2\\\\The\ sequence\ is \ (8)/(27),(4)/(9),(2)/(3),1,(3)/(2),...\\First\ term\ is (8)/(27)\\Common\ ratio\ is:(3)/(2)\\a) show\ that\ the\ series\ is\ a\ geometric\ progression:\\\\((4)/(9) )/((8)/(27) ) =(3)/(2) \\\\\\((2)/(3) )/((4)/(9) ) =(3)/(2) \\`

`b) Find\ the\ ninth\ term.\\\\1. for\ the\ sequence:\\\\u_0=(8)/(27) \\u_(n+1)=u_n*(3)/(2) \\\\n\geq 1:u_n=(8)/(27)*{((3)/(2) )}^(n-1)={((3)/(2) )}^(n-4)\\\\2. for\ the\ serie:\\\\\displaystyle\ \sum_(i=1)^(n)\ {((3)/(2) )}^(i-4)\\`