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3 years ago

Using a_n=a_1r^{n-1}, what is the 10th term in this geometric sequence?

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Mathematics

1 answer:

Whitepunk [10]3 years ago

5 0

$a_n=a_1.r^{n-1}\\
\\
a_{10}=(0.2).5^9\\
\\
a_{10}=(0.2).1,953,125=390,625$

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\begin{array}{lllll}
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% (a,b)
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3 years ago

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lord [1]

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Step-by-step explanation:

The given expression is :

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It can be solved as follows :

$\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}=\dfrac{a}{b(a-b)}+\dfrac{b}{a(b-a)}\\\\=\dfrac{a}{b(a-b)}+\dfrac{b}{-a(-b+a)}\\\\=\dfrac{1}{a-b}(\dfrac{a}{b}-\dfrac{b}{a})\\\\=\dfrac{a^2-b^2}{ab(a-b)}\\\\=\dfrac{(a-b)(a+b)}{ab(a-b)}\\\\=\dfrac{(a+b)}{ab}$

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Answer:

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3 years ago

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nexus9112 [7]

Answer:

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Step-by-step explanation:

let the numbers be x and y and x>y

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x=-9

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3 years ago