The sum of 4 consecutive numbers is 130, how do you find the 4 numbers?

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Answer 1 · 2018-05-02T03:31:46

Set up an equation where n = the first number and n +1 the second and n+2 the third and n+3 is the fourth.

$n + (n + 1) + (n + 2) + (n + 3) = 130$ $n + (n+1) + (n+2) + (n+3) = 130$

Combine like terms

$4 n + 6 = 130$ $4n + 6 = 130$ subtract 6 from both sides

$4 n + 6 - 6 = 130 - 6$ $4n + 6 - 6 = 130 -6$ which gives

$4 n = 124$ $4n = 124$ Divide both sides by 4

$4 \frac{n}{4} = \frac{124}{4}$ $4n/4 = 124/4$ so

$n = 31$ $n = 31$
$n + 1 = 32$ $n+1 = 32$
$n + 2 = 33$ $n +2 = 33$
$n + 3 = 34$ $n+3 = 34$