Question

The sum of three consecutive numbers in an AP is 24 and their product is 440. Find the numbers if the common difference of the AP is positive.

• A. 6, 8, 10
• B. 6, 9, 12
• C. 5, 8, 11
• D. 0, 8, 16

Answer 1

The correct option is C 5, 8, 11

Let the three consecutive numbers be (a - d), a, (a + d).
Therefore, (a - d) + a + (a + d) = 24.
a - d + a + a + d = 24.
3a = 24.
a = 8.
Given the product is 440.
Hence, (a - d)(a)(a + d) = 440.
$(a^2-d^2)a=440(64-d^2)8=44064-d^2=55d^2=9d=3(Sincedisgiventobepositive.)$
Therefore, the required numbers are (a - d), a, (a + d) = (8 - 3), 8, (8 + 3)
= 5, 8, 11