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The sum of 3 consecutive multiples of 4 is 120. Find the numbers.

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Answer:

The required numbers are: 36, 40, 44

Explanation:

Let first number = 4x

Second number = 4(x+1)

Third number = 4(x+2)

There sum = 120

We need to find the numbers

We can write:


4x+4(x+1)+4(x+2)=120\\4x+4x+4+4x+8=120\\12x+12=120\\12x=120-12\\12x=108\\x=(108)/(12)\\x=9

So, we get x = 9

Now, finding the numbers:

first number = 4x = 4(9) =36

Second number = 4(x+1) = 4(9+1) = 4(10) = 40

Third number = 4(x+2) = 4(9+2) = 4(11) = 44

So, The required numbers are: 36, 40, 44

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