Four consecutive integer numbers add to 2. what are the numbers?

The numbers are consecutive, meaning they are in arithmetic progression.

Sum of Arithmetic Progression = (n/2)[2a + (n- 1)d]

Where, n = total number of numbers and d = common difference between two successive numbers

Therefore n = 4 and d = 1 ( as the numbers are consecutive )

now (4/2)(2a + (4 - 1)1) = 2

i.e. 2(2a + 3) = 2

i.e. 4a + 6 = 2

i.e. 4a = - 4

i.e. a = - 1

Therefore nos are -1, -1+1 , -1+1+1 , -1+1+1+1

i.e. -1, 0 , 1 , 2

Hope this helps..!!

Thank you :)

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