Question: A sequence of four real numbers forms an arithmetic progression. If the sum of the first and last terms is 14 and the product of the second and third terms is 24, find the common difference.

Sat, 02 May 2026 11:14:37 +0000

**Why So Many Are Analyzing Arithmetic Progressions These Days**
In an era driven by pattern recognition and data-driven decisions, almost anyone interested in math, finance, architecture, or tech trends encounters word problems that blend logic and real-world relevance. This particular sequence puzzle—four real numbers in arithmetic progression with a fixed sum and product relationship—has gained quiet traction online. It reflects a growing curiosity in understanding how simple equations map to tangible systems, from investment growth models to spatial design. Users aren’t just seeking answers; they’re exploring how structured thinking can simplify complex systems. As mobile search habits lean toward quick yet insightful learning, this type of problem stands out in context-rich queries, offering both clarity and curiosity. **Why This Question Is Trending in the US Context**
Across the United States, interest in foundational math problems grows amid a broader focus on financial literacy, data interpretation, and problem-solving skills among students and professionals alike. This sequence challenge taps into that mindset—offering a relatable, low-stakes way to practice algebra while building analytical confidence. Many users engage with it not out of academic pressure, but because they’re drawn to puzzles that mirror real-life patterns, from tracking monthly income streams to modeling real estate returns. The problem’s accessibility and logical structure make it ideal for educational content in Discover, aligning with mobile-first habits of quick, meaningful comprehension. **How the Sequence Works: Breaking It Down Carefully**
In an arithmetic progression, the terms follow a steady, constant difference—the common difference, usually denoted as *d*. Let the first term be *a*. Then the four terms can be written as:
a, a + d, a + 2d, a + 3d
The sum of the first and last terms is:
a + (a + 3d) = 14 → 2a + 3d = 14
The second and third terms are:
a + d and a + 2d
Their product is:
(a + d)(a + 2d) = 24
Expanding this gives a quadratic in *a* and *d*
(a² + 3ad + 2d²) = 24
Combining both equations provides a clear path: use 2a = 14 – 3d → substitute into the product equation. Solving expands into a quadratic *ad² + bd + c = 0*, solvable via