Question

what is the smallest of four consecutive odd integers whose sum is 968?

Answer 1

Answer:

smallest of four consecutive odd integers whose sum is = 239

Step-by-step explanation:

Four consecutive odd integers are n, n+2, n+4, n+6

n+(n+2)+(n+4)+(n+6) = 968

             4n+12 = 968

                4n = 956

                 n = 239

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Answer 2

Answer:

239

Step-by-step explanation:

let smallest odd integer be x

since consecutive odd integers will differ by 2 (for ex. 1,3,5...)

The other three odd integers = (x+2), (x+4), (x+6)

Given that their sum=968,

x + x  + 2 + x + 4 + x + 6 = 968

4x + 12 = 968

4x = 968-12 = 956

x = 956/4 = 239

you could also do this by A.P. method

where first term = a = x

common difference = d = 2

number of terms = 4

sum = n/2 ( 2a + (n-1)d )

968 = 4/2 ( 2x + (4-1)2)

968 = 2 (2x + 3×2)

968/2 = 2x + 6

484 = 2x + 6

2x = 484 - 6 = 478

x= 478/2 = 239