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horsena [70]
3 years ago
6

What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?

Mathematics
1 answer:
Trava [24]3 years ago
5 0

Answer:

\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}

Step-by-step explanation:

The explicit formula for a geometric sequence:

a_n=a_1r^{n-1}

a_n - n-th term

a_1 - first term

r - common ratio

r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}

We have

a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...

The common ratio:

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}

<h2>It's not a geometric sequence.</h2>

If a_3=-\dfrac{1}{16} then the common ratio is r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}

Put to the explicit formula:

a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}

Put a_n=\dfrac{1}{1024} and solve for <em>n </em>:

-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

\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6

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Answer:

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With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

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S_{x1-x2 = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

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Critical Value = t_{\frac{\alpha }{2}, df = t_{0.05, df=13 = 1.771  { t-table }

So,

Margin of Error E =  t_{\frac{\alpha }{2}, df × [ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

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Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

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