Respuesta :
There are 8 terms of the series required to give the given sum
Step-by-step explanation:
The formula of the sum of n terms of a geometric series is
[tex]S_{n}=\frac{a(1-r^{n})}{1-r}[/tex] ,where
- a is the first term in the series
- r is the common ratio between each two consecutive terms
- [tex]r=\frac{U_{2}}{U_{1}}[/tex]
∵ The series is 125 + 25 + 5 + 1 + ............
∵ [tex]\frac{25}{125}=\frac{1}{5}[/tex]
∵ [tex]\frac{5}{25}=\frac{1}{5}[/tex]
- There is a common ratio between each two consecutive terms
∴ The series is a geometric series, where a = 125 and r = [tex]\frac{1}{5}[/tex]
∵ [tex]S_{n}=\frac{a(1-r^{n})}{1-r}[/tex]
∵ a = 125 , r = [tex]\frac{1}{5}[/tex] and [tex]S_{n}=\frac{97656}{625}[/tex]
- Substitute these values in the rule above
∴ [tex]\frac{97656}{625}=\frac{125[1-(\frac{1}{5})^{n}]}{1-\frac{1}{5}}[/tex]
∴ [tex]\frac{97656}{625}=\frac{125[1-(\frac{1}{5})^{n}]}{\frac{4}{5}}[/tex]
∵ [tex]\frac{125}{\frac{4}{5}}=\frac{625}{4}[/tex]
∴ [tex]\frac{97656}{625}=\frac{625[1-(\frac{1}{5})^{n}]}{4}[/tex]
- Divide both sides by [tex]\frac{625}{4}[/tex]
∴ [tex]0.99999744=1-(\frac{1}{5})^{n}[/tex]
- Subtract 1 from both sides
∴ [tex]-\frac{1}{390625}=-(\frac{1}{5})^{n}[/tex]
- Multiply both sides by -1
∴ [tex]\frac{1}{390625}=(\frac{1}{5})^{n}[/tex]
- Insert ㏑ in both sides
∴ ㏑ [tex](\frac{1}{390625})[/tex] = ㏑ [tex](\frac{1}{5})^{n}[/tex]
- Remember ㏑ [tex](a)^{n}[/tex] = n ㏑(a)
∴ ㏑ [tex](\frac{1}{390625})[/tex] = n ㏑ [tex](\frac{1}{5})[/tex]
- Divide both sides by ㏑ [tex](\frac{1}{5})[/tex] to find n
∴ n = 8
There are 8 terms of the series required to give the given sum
Learn more:
You can learn more about the geometric series in brainly.com/question/1522572
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