The average of three consecutive integers is 20. What is the largest of these integers?

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Q: The average of three consecutive integers is 20. What is the largest of these integers?

Solution: Let the integers be x, x+1, x+2. Average = (x + (x+1) + (x+2)) / 3 = 20. Solving gives x = 19, largest integer = 21.

Steps: 10

Step 1: Understand that we are looking for three consecutive integers. Let's call the first integer 'x'.
Step 2: The next two consecutive integers can be expressed as 'x + 1' and 'x + 2'.
Step 3: To find the average of these three integers, we add them together: x + (x + 1) + (x + 2).
Step 4: Simplify the sum: x + x + 1 + x + 2 = 3x + 3.
Step 5: Now, we divide the total by 3 to find the average: (3x + 3) / 3.
Step 6: Set the average equal to 20: (3x + 3) / 3 = 20.
Step 7: Multiply both sides by 3 to eliminate the fraction: 3x + 3 = 60.
Step 8: Subtract 3 from both sides: 3x = 57.
Step 9: Divide by 3 to solve for x: x = 19.
Step 10: Now, find the largest integer, which is x + 2: 19 + 2 = 21.