The sum of three consecutive odd integers is 76 less then seven times the middle number. The three integers are 17, 19 and 21 respeectively
Solution:
Since each consecutive odd integer is separated by a difference of 2
Let "n" be the first integer
n + 2 be the second integer
n + 4 be the third integer
Given that the sum of three consecutive odd integers is 76 less then seven times the middle number
Which means,
The sum of ( n, n + 2, n + 4) is equal to 76 less than seven times the middle number ( 7(n + 2))
That is,
n + n + 2 + n + 4 = 7(n + 2) - 76
3n + 6 = 7n + 14 - 76
4n = 68
n = 17
So we get:
First integer = n = 17
Second integer = n + 2 = 17 + 2 = 19
Third integer = n + 4 = 17 + 4 = 21
Thus the three consecutive odd integers are 17, 19 and 21 respeectively