MATH SOLVE

2 months ago

Q:
# What is the explicit rule for this geometric sequence? 4,45,425,4125,…

Accepted Solution

A:

You did not write the question correctly but I will try to answer the question according to 2 possible interpretations of your question.

Case 1:

Given the geometric sequence:

[tex]4,\ \frac{4}{5} ,\ \frac{4}{25} ,\ \frac{4}{125} ,\ .\ .\ .[/tex]

Common ratio is given by [tex]r= \frac{2nd\ term}{1st\ term} = \frac{ \frac{4}{5} }{4} = \frac{1}{5} [/tex]

The explicit rule of a geometric sequence is given by [tex]T_n=ar^{n-1}[/tex]

Therefore, the explicit rule of the given sequence is [tex]T_n=4\left( \frac{1}{5} \right)^{n-1}\ or\ 20\left( \frac{1}{5} \right)^n[/tex]

Case 2:

Given the geometric sequence:

[tex]4,\ 4\cdot5,\ 4\cdot25,\ 4\cdot125,\ .\ .\ .[/tex]

Common ratio is given by [tex]r= \frac{2nd\ term}{1st\ term} = \frac{4\cdot5}{4} = 5 [/tex]

The explicit rule of a geometric sequence is given by [tex]T_n=ar^{n-1}[/tex]

Therefore, the explicit rule of the given sequence is [tex]T_n=4(5)^{n-1}\ or\ \frac{4}{5}(5)^n[/tex]

Case 1:

Given the geometric sequence:

[tex]4,\ \frac{4}{5} ,\ \frac{4}{25} ,\ \frac{4}{125} ,\ .\ .\ .[/tex]

Common ratio is given by [tex]r= \frac{2nd\ term}{1st\ term} = \frac{ \frac{4}{5} }{4} = \frac{1}{5} [/tex]

The explicit rule of a geometric sequence is given by [tex]T_n=ar^{n-1}[/tex]

Therefore, the explicit rule of the given sequence is [tex]T_n=4\left( \frac{1}{5} \right)^{n-1}\ or\ 20\left( \frac{1}{5} \right)^n[/tex]

Case 2:

Given the geometric sequence:

[tex]4,\ 4\cdot5,\ 4\cdot25,\ 4\cdot125,\ .\ .\ .[/tex]

Common ratio is given by [tex]r= \frac{2nd\ term}{1st\ term} = \frac{4\cdot5}{4} = 5 [/tex]

The explicit rule of a geometric sequence is given by [tex]T_n=ar^{n-1}[/tex]

Therefore, the explicit rule of the given sequence is [tex]T_n=4(5)^{n-1}\ or\ \frac{4}{5}(5)^n[/tex]