Answer:
[tex]a_n =80(0.5)^{n-1}[/tex]
Step-by-step explanation:
we know that
The explicit formula for a geometric sequence is written as:
[tex]a_n = a_1(r)^{n-1}[/tex]
where
a_1 is the first term
r is the common ratio
we have the sequence
[tex]80,40,20,10,..[/tex]
so
[tex]a_1=80\\a_2=40\\a_3=20\\a_4=10[/tex]
[tex]\frac{a_2}{a_1}=\frac{40}{80}=0.5[/tex]
[tex]\frac{a_3}{a_2}=\frac{20}{40}=0.5[/tex]
[tex]\frac{a_4}{a_3}=\frac{10}{20}=0.5[/tex]
therefore
[tex]r=0.5[/tex]
substitute
[tex]a_n =80(0.5)^{n-1}[/tex]
Answer:
[tex]\displaystyle a_n=80\times \left(\frac{1}{2}\right)^{n-1}[/tex]
Step-by-step explanation:
Geometric Sequence
It's a particular type of sequence where the next term is computed as the previous one times a constant number called common ration. The general formula for a geometric sequence is
[tex]a_n=a_1\cdot r^{n-1}[/tex]
where n is the number of the term
We have to help Divya and Miguel to find an explicit formula for the sequence 80,40,20,10,..
The first term is
[tex]a_1=80[/tex]
We need to compute the common ratio, knowing that
[tex]a_2=40[/tex]
From the general formula
[tex]a_n=a_1\cdot r^{n-1}[/tex]
we solve for r
[tex]\displaystyle r=\sqrt[n-1]{\frac{a_n}{a_1}}[/tex]
[tex]\displaystyle r=\sqrt[2-1]{\frac{40}{80}}[/tex]
[tex]\displaystyle r=\frac{1}{2}[/tex]
Let's test further to make sure the sequence has a common ratio
[tex]\displaystyle a_3=a_1.r^2=80\times \left(\frac{1}{2}\right)^2[/tex]
[tex]a_3=20[/tex]
Similarly we find
[tex]a_4=10[/tex]
The required explicit formula is
[tex]\boxed{\displaystyle a_n=80\times \left(\frac{1}{2}\right)^{n-1}}[/tex]
