if the sum of 3 consecutive numbers is 123, what are the 3 numbers?

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  • John P
    Lv 7
    4 weeks ago

    If you think about it you can do it in your head and on a scrap of paper. 3 consecutive numbers, so each must be about one third of 120. That's 40. Add on a bit, one more each time, and you are there.  No real need for algebra.

  • 1 month ago

    n + (n + 1) + (n + 2) = 123

    3n + 3 = 123

     3n = 123 - 3

    3n = 120

    n = 40

    n + 1 = 40 + 1 = 41

    n + 2 = 40 + 2 = 42

    Therefore the 3 numbers are: 40,41,42..

  • 1 month ago

    x + (x+1) + (x+2) = 123

    3x + 3 = 123

    3x = 120

    x = 40

    4 numbers are 40, 40+1 and 40+2

  • 1 month ago

    40 , 41 , 42 ..

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  • Phil
    Lv 6
    1 month ago

    The three consecutive numbers are 40, 41 and 42.

  • 1 month ago

    Hi

    Marianna

    Let,

    The sum of 3 consecutive numbers is 123

    numbers are (n-1), n, (n+1) 

    sum = 3n 

    so,

     n = 123/3 

            

             = 41

    Hence,the final answer is  40,41, 42

  • 1 month ago

    numbers are (n-1), n, (n+1) sum = 3n so n = 123/3 = 41

    hence 40,41, 42

  • 1 month ago

    There's no such thing as "consecutive numbers."

    Did you mean consecutive integers? Consecutive odd numbers? Consecutive even numbers? Consecutive perfect squares? Please clarify.

  • 1 month ago

    The 3 consecutive numbers are 40, 41, and 42.

  • 1 month ago

    A shortcut to figuring these out is to realize the middle number will be the average.

    Divide by 3 and that's the middle number.

    123/3 = 41

    Answer:

    40, 41, 42

    You can see why this works algebraically if you let x represent the *middle* integer. The prior integer will be x-1 and the subsequent integer will be x+1.

    x-1 + x + x+1 = 123

    3x = 123

    x = 123/3

    x = 41 <-- middle integer

    Answer:

    40, 41, 42

    A more "typical" way to do this, though, is to let x represent the smallest integer.

    Let x be the smallest integer

    Let x+1 be the middle integer

    Let x+2 be the largest integer

    x + x+1 + x+2 = 123

    3x + 3 = 123

    3x = 120

    x = 120/3

    x = 40 <-- smallest integer

    Answer:40, 41, 42

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