Three numbers are equally spaced (they form an arithmetic progression). If…
2025
Three numbers are equally spaced (they form an arithmetic progression). If their average is 135 and the common difference between consecutive numbers is 25, find the lowest number.
- A.
90
- B.
110
- C.
160
- D.
180
Attempted by 83 students.
Show answer & explanation
Correct answer: B
Answer: 110 (the lowest number).
Reasoning: Because the three numbers are equally spaced with a common difference of 25, write them as y - 25, y, and y + 25, where y is the middle number.
The average of an evenly spaced set equals its middle number, so the average equation is (y - 25 + y + y + 25) / 3 = 135.
Simplify the sum: (3y) / 3 = 135, so y = 135 (the middle number equals the average).
Find the numbers: y - 25 = 110, y = 135, y + 25 = 160.
Therefore the lowest number is 110. Check: 110 + 135 + 160 = 405 and 405 / 3 = 135.