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3 years ago

What is the recursiveformula for this geometric sequence? 4,-12,36,108

Mathematics

1 answer:

Vlad1618 [11]3 years ago

8 0

Answer:

a[1] = 4

a[n] = -3·a[n-1]

Step-by-step explanation:

The sequence given is not a geometric sequence, since the ratios of terms are -3, -3, 3 -- not a constant.

If we assume that the last given term is supposed to be -108, then the common ratio is -3 and each term is -3 times the previous one. That is expressed in a recursive formula as ...

a[1] = 4 . . . . . . . . . . . first term is 4

a[n] = -3·a[n-1] . . . . . each successive term is -3 times the previous one

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2 years ago

Read 2 more answers

Write 2.71x10-3 as an ordinary number.

Katyanochek1 [597]

Step-by-step explanation:

${2.71 \times 10}^{ - 3}$

This can be rewritten as

$2.71 \times \frac{1}{10^{3} }$

If we will solve this way,

$\frac{2.71}{1000} = 0.00271$

But since the denominator is just in terms of 1, we can just simply get the (-3). Note that we have to take the negative into consideration.

From the original position of the decimal point, move the decimal point 3 times TO THE LEFT. Therefore, from 2.71 we move the decimal point 3 times to the left, 0.00271.

If the exponent has a positive value, move TO THE RIGHT.

For example,

$2.71 \times {10}^{3} = 2.71 \times 1000 \\ = 2710$

Note that the exponent has a positive value. Therefore, we have to move the decimal point 3 times TO THE RIGHT. Producing 2710 as the answer.

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2 years ago

nikitadnepr [17]

The answer is all real numbers

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2 years ago