Respuesta :
[tex]\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------[/tex]
[tex]\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ r=-\frac{1}{4}\\ a_1=-4096\\ n=6 \end{cases} \\\\\\ a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5[/tex]
[tex]\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ r=-\frac{1}{4}\\ a_1=-4096\\ n=6 \end{cases} \\\\\\ a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5[/tex]
Answer:
The sixth term of the geometric sequence is 4.
Step-by-step explanation:
Given : Geometric sequence term are [tex]a_1=-4096[/tex] and [tex]a_4=64[/tex]
To find : What is the 6th term of geometric sequence ?
Solution :
[tex]a_1=-4096[/tex]
We know, fourth term is [tex]a_4=ar^3[/tex]
[tex]a_4=64[/tex]
So, [tex]64=(-4096)r^3[/tex]
[tex]\frac{64}{(-4096)}=r^3[/tex]
[tex]-0.015625=r^3[/tex]
[tex]r=(-0.015625)^{\frac{1}{3}}[/tex]
The sixth term is [tex]a_6=ar^5[/tex]
Substitute,
[tex]a_6=(-4096)((-0.015625)^{\frac{1}{3}})^5[/tex]
[tex]a_6=-4096((-0.015625)^{\frac{5}{3}})[/tex]
[tex]a_6=-4096(-0.0009765625)[/tex]
[tex]a_6=4[/tex]
Therefore, The sixth term of the geometric sequence is 4.
