Q: Find three consecutive multiples of 15 such that twice their sum is 360?

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18, 20 and 22

They are 14, 16 and 18.

The numbers must total 888/8 ie 111. The middle number is 1/3 of 111 ie 37 so the three numbers are 36, 37 and 38 and multiples are 288, 296 and 304.

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

-10 and -11.

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91

18, 20 and 22

Let the second of the three consecutive multiples of 6 be 6n Then the first is 6n - 6 and the last is 6n + 6; and: (6n - 6) + 6n + (6n + 6) = 666 → 18n = 666 → n = 37 → the consecutive multiples of 6 which sum to 666 are 216, 222, 228

They are 14, 16 and 18.

The numbers are 14, 16 and 18.

5872345098234783904672083946728390752430689723409687298290843 theres your answer

8A + (8A+8) + (8A+16) = 888 24A + 24 = 888 ===> 24A = 864 ===> A = 36 The multiples are 8A, 8A+8, 8A+16 = 288, 296, 304. Sum = 888.

72, 81, 90

The numbers must total 888/8 ie 111. The middle number is 1/3 of 111 ie 37 so the three numbers are 36, 37 and 38 and multiples are 288, 296 and 304.

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2, 3, 4, 5 and 6 are multiples of 2, 3, 4, 5 and 6.

The three integers, since they are consecutive, can be listed as a, a+1, and a+2. Twice the first is 2a. Three times the third is 3(a+2) = 3a+6. First make a formula of the information given: 2a+(3a+6)= -24 Next, solve the formula: 5a + 6 = -24 Subtract 6 from each side. 5a = -30 Divide each side by 5. a = -6 The three consecutive numbers are -6, -5, and -4.