The sum of three consecutive integers is 192. what are the integers

63, 64, 65

Step-by-step explanation:

Consecutive integers have a difference of 1 between them

let the 3 integers be n, n + 1 and n + 2 , then their sum equals 192

n + n + 1 + n + 2 = 192

3n + 3 = 192 ( subtract 3 from both sides )

3n = 189 ( divide both sides by 3 )

n = 63 ⇒ n + 1 = 64 and n + 2 = 65

the 3 integers are 63, 64 and 65

63, 64, and 65

Step-by-step explanation: This problem states that the sum of three consecutive integers is 192 and it asks us to find the integers. We can represent our three consecutive integers as followed.

X ⇒ first integer

X + 1 ⇒ second integer

X + 2 ⇒ third integer

Since the sum of our three consecutive integers is 192, we can set up an equation to represent this.

X + X + 1 + X + 2 = 192

We can simplify on the left side of our equation.

3x + 3 = 192

      -3     -3   ← subtract 3 from both sides of the equation

3x = 189

÷3     ÷3   ← divide both sides by 3

 X = 63

X ⇒ first integer = 63

X + 1 ⇒ second integer = 64

X + 2 ⇒ third integer = 65

Finally, it's always a good idea to go back and check to see if your answer is correct. To see if your answer is correct, we will add our consecutive integers and they should have a sum of 192.

63 + 64 + 65 = 192

Therefore, our three consecutive integers are 63, 64, and 65.

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation:

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58