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2 years ago

What is the eighth term of the geometric sequence 6, 18, 54

Mathematics

1 answer:

Rainbow [258]2 years ago

4 0

Answer:

= 4374.

Step-by-step explanation:

it is important to understand the pattern hidden in such problem.Let’s give it a try 2, 6, 18, 54 and so on.It can be written as 2, 3*2, 9*2, 27*2 and so on.

This can be further written as2 (1, 3, 9, 27, and so on) as 2 is common in every term.Now if you see the chain 1,3,9, 27 and so….you will see a pattern hidden i.e. 3=1*3 ,9=1*3*3, 27=1*3*3*3 now 27 is the 4th term consist of three 3. So 8th term would consist of seven 3. 8th would be 8th term = 1*3*3*3*3*3*3*3 = 2187 Hence the 8th term for the series 2,6,18,54 would be= 2*2187 = 4374.

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2x = 48, 3x = 72, 4x = 96, 6x = 144

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Let H be the set of all polynomials having degree at most 4 and rational coefficients. Determine whether H is a vector space. If

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Yes. It is a vector space over the field of rational numbers $\mathbb{Q}$

Step-by-step explanation:

An element $p$ of the set $H$ has the form

$p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}$

where $a_{0},a_{1},a_{2},a_{3},a_{4}$ are rational coefficients.

The operations of addition and scalar multiplication are defined as follows:

$p(x)+q(x)=(a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+x_{4}x^4)+(b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4})=(a_{0}+b_{0})+(a_{1}+b_{1})x+(a_{2}+b_{2})x^{2}+(a_{3}+b_{3})x^{3}+(a_{4}+b_{4})x^{4}$

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The properties that $H$ , together the operations of vector addition and scalar multiplication, must satisfy are:

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Distributive properties.

This is not difficult with the definitions given. The most important part is to show that $H$ has a additive identity, which is the zero polynomial, that is closed under vector addition and scalar multiplication. This last properties comes from the fact that $\mathbb{Q}$ is a field, then it is closed under sum and multiplication.

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