Q: The sum of 3 consecutive integers is twice the first integer?

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They are 14, 16 and 18.

9 and 10 9 + 2(10) = 29

9, 11, and 139 + 2(11) + 3(13) = 9 + 22 + 39 = 70

18, 20 and 22

55 and 57

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They are 14, 16 and 18.

The numbers are 14, 16 and 18.

9 and 10 9 + 2(10) = 29

9, 11, and 139 + 2(11) + 3(13) = 9 + 22 + 39 = 70

The answer would be 10 12 and 14... 14 x 3 = 42 and 2(10 + 12) = 44. So the product of the largest integer and three is two less than twice the sum of the lower integers.

Suppose the smallest integer is A. The next two even numbers are A+2 and A+4. Using the information supplied we can form an equation: 2A - 14 = A + A+2 + A+4 Rearranging: 2A - 14 = 3A + 6 -20 = A So the three integers are -20, -18 and -16.

18, 20 and 22

The integers are 14 and 7.

no one wants to know the answer. its freaking math

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If the largest integer is subtracted from four times the smallest, the result is 4 more than twice the middle integer. Let the smallest integer be x, then the others are x + 2 and x+ 4. Therefore 4x - (x + 4) = 2 (x + 2 ) + 4 Expanding, we get 4x -x -4 = 2x + 4 + 4 Gathering terms: x = 12 Thus the three integers are 12, 14 and 16.

-10 and -11.