Answer:
[tex]\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}[/tex]
Step-by-step explanation:
The explicit formula for a geometric sequence:
[tex]a_n=a_1r^{n-1}[/tex]
[tex]a_n[/tex] - n-th term
[tex]a_1[/tex] - first term
[tex]r[/tex] - common ratio
[tex]r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}[/tex]
We have
[tex]a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...[/tex]
The common ratio:
[tex]r=\dfrac{\frac{1}{4}}{-1}=-\dfrac{1}{4}\\\\r=\dfrac{-\frac{1}{6}}{\frac{1}{4}}=-\dfrac{1}{6}\cdot\dfrac{4}{1}=-\dfrac{2}{3}\neq-\dfrac{1}{4}[/tex]
If [tex]a_3=-\dfrac{1}{16}[/tex] then the common ratio is [tex]r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}[/tex]
Put to the explicit formula:
[tex]a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}[/tex]
Put [tex]a_n=\dfrac{1}{1024}[/tex] and solve for n :
[tex]-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1[/tex]
[tex]\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6[/tex]
