Answer:
The sum of the first six terms is
[tex]32-8+2-0.5+\frac{1}{8}-\frac{1}{32}=25.59[/tex]
Step-by-step explanation:
Given series is 32-8+2-0.5+...
We may write [tex]{\{32,-8,2,-0.5,...}\}[/tex]
Let [tex]a_{1}=32,a_{2}=-8,a_{3}=2,a_{4}=-05,...[/tex]
Common ratio [tex]r=\frac{a_{2}}{a_{1}}[/tex]
[tex]r=\frac{-8}{32}[/tex]
[tex]r=\frac{-1}{4}[/tex]
[tex]r=\frac{a_{3}}{a_{2}}[/tex]
[tex]r=\frac{2}{-8}[/tex]
[tex]r=\frac{-1}{4}[/tex]
Therefore the common ratio is [tex]r=\frac{-1}{4}[/tex]
Therefore given sequence is of the form of Geometric sequence
The nth term of the geometric sequence is
[tex]a_{n}=ar^{n-1}[/tex]
First to find the 5th and 6th term
That is substitute n=5 and n=6 ,a=32 and [tex]r=\frac{-1}{4}[/tex] in above equation we get
[tex]a_{5}=32(\frac{-1}{4})^{5-1}[/tex]
[tex]a_{5}=32(\frac{-1}{4})^{4}[/tex]
[tex]a_{5}=32[(\frac{-1}{4})\times (\frac{-1}{4})\times (\frac{-1}{4})\times (\frac{-1}{4})][/tex]
[tex]=\frac{1}{8}[/tex]
Therefore [tex]a_{5}=\frac{1}{8}[/tex]
[tex]a_{6}=32(\frac{-1}{4})^{6-1}[/tex]
[tex]a_{6}=32(\frac{-1}{4})^{5}[/tex]
[tex]a_{6}=32[(\frac{-1}{4})\times (\frac{-1}{4})\times (\frac{-1}{4})\times (\frac{-1}{4})\times (\frac{-1}{4})][/tex]
[tex]=\frac{-1}{32}[/tex]
Therefore [tex]a_{6}=-\frac{1}{32}[/tex]
Therefore the sum of the first six terms is
[tex]32-8+2-0.5+\frac{1}{8}-\frac{1}{32}[/tex]
[tex]=25.5+\frac{3}{32}[/tex]
[tex]=\frac{819}{32}[/tex]
[tex]=25.59[/tex]
Therefore the sum of the first six terms is
[tex]32-8+2-0.5+\frac{1}{8}-\frac{1}{32}=25.59[/tex]
