Seven consecutive integers are added. the sum is 6 times the greatest of the seven integers .what is the smallest integer

the smallest integer is 15

note that consecutive integers have a difference of 1 between them

let the smallest integer be n

then : n, n+ 1, n + 2, n + 3, n + 4, n + 5, n + 6

are seven consecutive integers with n + 6 being the largest, thus summing gives

7n + 21 = 6(6 + n)

7n + 21 = 36 + 6n ( distributing right side )

7n - 6n + 21 = 36 ( subtracting 6n from both sides )

n + 21 = 36 ⇒ n = 36 - 21 = 15

the smallest integer is 15

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

my guess would be b and d, but i didn’t actually get out the pen and paper. lol