Split 124 into four terms in AP with a product condition: Divide 124 into four parts in arithmetic progression such that the product of the 1st and 4th is 128 less than the product of the 2nd and 3rd. Identify the four parts.

Introduction / Context:
Let four AP terms be a−3d, a−d, a+d, a+3d. Their sum fixes a, and a product relation ties down d. This is a classic AP-parameterization technique for constrained partition problems.

Given Data / Assumptions:

• Total sum = 124.
• AP terms: T1 = a−3d, T2 = a−d, T3 = a+d, T4 = a+3d.
• T1*T4 is 128 less than T2*T3.

Concept / Approach:
Use sum to get a, then convert the product condition into an equation in d via the identities (a−3d)(a+3d) = a^2 − 9d^2 and (a−d)(a+d) = a^2 − d^2.

Step-by-Step Solution:
Sum: (a−3d)+(a−d)+(a+d)+(a+3d) = 4a = 124 ⇒ a = 31Product condition: (a^2 − 9d^2) = (a^2 − d^2) − 128⇒ −9d^2 = −d^2 − 128 ⇒ 8d^2 = 128 ⇒ d^2 = 16 ⇒ d = ±4Terms: 31−12 = 19, 31−4 = 27, 31+4 = 35, 31+12 = 43

Verification / Alternative check:
Products: 19*43 = 817; 27*35 = 945; indeed 817 = 945 − 128.

Why Other Options Are Wrong:
Other sets fail either the AP structure with correct center 31 or the exact 128 product difference.

Common Pitfalls:
Using sequential integers instead of AP form; mixing up which product is smaller.

Final Answer:
19, 27, 35, 43