How do you find four consecutive integers such that the sum of the two largest subtracted from twice the sum of the two smallest is 15?

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How do you find four consecutive integers such that the sum of the two largest subtracted from twice the sum of the two smallest is 15?

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Answer 1 · 2015-06-18T23:10:14

Substitute $n$ $n$ , $n + 1$ $n+1$ , $n + 2$ $n+2$ and $n + 3$ $n+3$ into the description to get an equation in $n$ $n$ and solve it to find $n = 9$ $n=9$ .

Hence the integers are: $9$ $9$ , $10$ $10$ , $11$ $11$ and $12$ $12$ .

Explanation:

Let the four integers be $n$ $n$ , $n + 1$ $n+1$ , $n + 2$ $n+2$ and $n + 3$ $n+3$

The sum of the two largest integers is:

$(n + 2) + (n + 3) = 2 n + 5$ $(n+2)+(n+3) = 2n+5$

Twice the sum of the two smallest integers is:

$2 (n + (n + 1)) = 2 (2 n + 1) = 4 n + 2$ $2(n + (n+1)) = 2(2n+1) = 4n+2$

We are given:

$15 = 2 (n + (n + 1)) - ((n + 2) + (n + 3))$ $15 = 2(n + (n+1)) - ((n+2)+(n+3))$

$= (4 n + 2) - (2 n + 5) = 2 n - 3$ $= (4n+2) - (2n+5) = 2n-3$

Add $3$ $3$ to both ends to get: $2 n = 18$ $2n=18$

Divide both sides by $2$ $2$ to get $n = 9$ $n=9$

So the four integers are: $9$ $9$ , $10$ $10$ , $11$ $11$ , $12$ $12$

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How do you find four consecutive integers such that the sum of the two largest subtracted from twice the sum of the two smallest is 15?

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