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3 years ago

Which sequences are geometric? Check all that apply.

Mathematics

1 answer:

NikAS [45]3 years ago

6 0

Step-by-step explanation:

A geometric sequence is when we multiply one number by a set rate to obtain the next number.

For the first one, we multiply 5 by 2 to get 10, 10 by 2 to get 20, and so on. This is geometric

Next, we multiply 3 by 4 to get 12, 12 by 4 to get 48, and so on. This is geometric

Next, we multiply 3 by 5 to get 15, 15 by 5 to get 75, and so on. This is geometric

Next, we multiply 8 by 15/8 to get 15, and 15 by 5 to get 75. 15/8 and 5 are not the same so this is not geometric

Next, we multiply 14 by 21/14 to get 21, and 21 by 28/21 to get 28. 21/14 and 28/21 are not the same so this is not geometric

Next, we multiply 17 by 20/17 to get 20, and 20 by 23/20 to get 23. 20/17 and 23/20 are not the same so this is not geometric.

Finally, we multiply 2 by 5 to get 10, 10 by 5 to get 50, and so on. This is geometric

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4 years ago

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gulaghasi [49]

the way I get the subsequent term, nevermind the exponents, the exponents part is easy, since one is decreasing and another is increasing, but the coefficient, to get it, what I usually do is.

multiply the current coefficient by the exponent of the first-term, and divide that by the exponent of the second-term + 1.

so if my current expanded term is say 7a³b⁴, to get the next coefficient, what I do is (7*3)/5 <----- notice, current coefficient times 3 divided by 4+1.

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$\bf ~~~~~~~~\textit{binomial theorem expansion} \\\\ \qquad \qquad (1+ax)^n\implies \begin{array}{llll} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(1)^n(ax)^0\\\\ 2&+\frac{(1)(n)}{1}\to n&(1)^{n-1}(ax)^1\\\\ 3&+\frac{n\cdot (n-1)}{2}&(1)^{n-2}(ax)^2 \end{array}$

so, following that to get the next coefficient, we get those equivalents as you see there for the 2nd and 3rd terms.

so then, we know that the expanded 2nd term is 24x therefore

$\bf n(1)^{n-1}(ax)1 = 24x\implies n(1)(ax)=24x\implies nax=24x\implies n=\cfrac{24}{a}$

we also know that the expanded 3rd term is 240x², therefore we can say that

$\bf \cfrac{n(n-1)}{2}~~(1)^{n-2}(ax)^2 = 240x^2\implies \cfrac{n(n-1)}{2}(1)(a^2x^2) = 240x^2 \\\\\\ \cfrac{(n^2-n)(a^2x^2)}{2}=240x^2\implies \cfrac{(n^2-n)(a^2)}{2}=\cfrac{240x^2}{x^2}\implies \cfrac{a^2n^2-a^2n}{2}=240 \\\\\\ a^2n^2-a^2n=480$

but but but, we know what "n" equals to, recall above, so let's do some quick substitution

$\bf a^2n^2-a^2n=480\qquad \boxed{n=\cfrac{24}{a}}\qquad a^2\left( \cfrac{24}{a} \right)^2-a^2\left( \cfrac{24}{a} \right)=480 \\\\\\ a^2\cdot \cfrac{24^2}{a^2}-24a=480\implies 24^2-24a=480\implies 576-24a=480 \\\\\\ -24a=-96\implies a=\cfrac{-96}{-24}\implies \blacktriangleright a = 4\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ n=\cfrac{24}{a}\implies n=\cfrac{24}{4}\implies \blacktriangleright n=6 \blacktriangleleft$

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Sorry if I wasn't specific enough, ask if you want more help. I am sure this is the right answer

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