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2 years ago

A sequence is given by rule 4n-5 generate the first 5 terms of the sequence

Mathematics

2 answers:

Illusion [34]2 years ago

7 0

So the answer is the following
-1, 3, 7, 15, 19

gtnhenbr [62]2 years ago

4 0

Answer:

- 1, 3, 7, 11, 15

Step-by-step explanation:

Substitute n = 1, 2, 3, 4, 5 into the rule, that is

a₁ = 4(1) - 5 = 4 - 5 = - 1

a₂ = 4(2) - 5 = 8 - 5 = 3

a₃ = 4(3) - 5 = 12 - 5 = 7

a₄ = 4(4) - 5 = 16 - 5 = 11

a₅ = 4(5) - 5 = 20 - 5 = 15

The first 5 terms are - 1, 3, 7, 11, 15

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Note: As you may have unintentionally missed to add the different answers, based on which we had to check who solved correctly between Tamara and Clyda's work.

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Answer:

We conclude that whoever (between Tamara and Clyda's work) may have got the answer as $x^2+5x-2$ after dividing $2x^4\:+\:7x^3\:-\:18x^2\:+\:11x\:-\:2$ by $2x^2\:-\:3x\:+\:1$ , would be the one who solved it correctly.

Step-by-step explanation:

Considering the expression

$2x^4\:+\:7x^3\:-\:18x^2\:+\:11x\:-\:2$

Lets divide the expression by $2x^2\:-\:3x\:+\:1$

Solution Steps:

$\frac{2x^4+7x^3-18x^2+11x-2}{2x^2-3x+1}$

Factorizing

$2x^4+7x^3-18x^2+11x-2:\quad \left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)$

$\frac{\left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)}{2x^2-3x+1}$

Factorizing

$2x^2-3x+1:\quad \left(2x-1\right)\left(x-1\right)$

$\frac{\left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)}{\left(2x-1\right)\left(x-1\right)}$

$\mathrm{Cancel\:}\frac{\left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)}{\left(2x-1\right)\left(x-1\right)}:\quad x^2+5x-2$

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Keywords: expression, division

Learn more about expression division from brainly.com/question/1575482

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A sequence is given by rule 4n-5 generate the first 5 terms of the sequence​

A sequence is given by rule 4n-5 generate the first 5 terms of the sequence