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2 years ago

FInd three consecutive even integers such that three times the smallest is four more than twice the largest

Mathematics

1 answer:

Oksanka [162]2 years ago

4 0

n, n + 2, n + 4 - three consecutive even integers

3n = 2(n + 4) + 4 |use distributive property

3n = (2)(n) + (2)(4) + 4

3n = 2n + 8 + 4

3n = 2n + 12 |subtract 2n from both sides

n = 12

n + 2 = 12 + 2 = 14

n + 4 = 12 + 4 = 16

Answer: 12, 14, 16.

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Answer: The fourth term is -102

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Explanation:

The term after the nth term is generated by this rule $a_{n+1} = -4(a_n) + 2$ which means that we first
Step 1) multiply the nth term ( $a_n$ ) by -4
Step 2) Add the result of step 1 to the value 2 to get the next term in the sequence

Let's follow those steps above to generate the first four terms

The first term is $a_1 = 2$ . In short, the first term is 2

The second term is...
$a_{n+1} = -4(a_n) + 2$
$a_{1+1} = -4(a_1) + 2$
$a_{2} = -4(2) + 2$
$a_{2} = -8 + 2$
$a_{2} = -6$
So the second term is -6

The third term is...
$a_{n+1} = -4(a_n) + 2$
$a_{2+1} = -4(a_2) + 2$
$a_{3} = -4(-6) + 2$
$a_{3} = 24 + 2$
$a_{3} = 26$
The third term is 26

Finally, the fourth term is...
$a_{n+1} = -4(a_n) + 2$
$a_{3+1} = -4(a_3) + 2$
$a_{4} = -4(26) + 2$
$a_{4} = -104 + 2$
$a_{4} = -102$
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