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Enter the finite geometric series from its given description, and then enter its sum.
A geometric series that starts with 4, ends with -12,500, and has a common ratio of -5.
The geometric series is _____
The sum of the geometric series is_____

Sagot :

semsee45

Answer:

[tex]a_n= 4(-5)^{(n-1)}[/tex]

-10416

Step-by-step explanation:

a = 4

r = -5

geometric series:  [tex]a_n=ar^{(n-1)} = 4(-5)^{(n-1)}[/tex]

Determine which term = -12500:

[tex]a_n=-12500[/tex]

[tex]4(-5)^{(n-1)}=-12500[/tex]

[tex](-5)^{(n-1)} = -3125[/tex]

[tex](5)^{(n-1)} = 3125[/tex]

[tex](n-1)ln(5) = ln(3125)[/tex]

[tex]n-1 = ln(3125)/ln(5) = 5[/tex]

[tex]n = 5 + 1 = 6[/tex]

[tex]a_6=-12500[/tex]

Using geometric sum series formula: [tex]S_n=\frac{a(1-r^n)}{1-r}[/tex]

Therefore, [tex]S_6=\frac{4(1-(-5)^6)}{1-(-5)} =\frac{-62496}{6} =-10416[/tex]

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