In the 1982 MAA Annual High School Mathematics Examination (Problem #8), students were presented with:

"By definition $r! = r(r - 1)\cdots 1$ and $\binom{j}{k} = \frac{j!}{k!(j - k)!}$. If $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ form an arithmetic progression with $n > 3$, then $n$ equals: (A) 5 (B) 7 (C) 9 (D) 11 (E) 12."

A sequence of three terms $A, B, C$ forms an arithmetic progression if the step from $A$ to $B$ equals the step from $B$ to $C$: $$B - A = C - B \iff 2B = A + C$$ Recall the algebraic formulas for these combinations: $$\binom{n}{1} = n, \qquad \binom{n}{2} = \frac{n(n - 1)}{2}, \qquad \binom{n}{3} = \frac{n(n - 1)(n - 2)}{6}$$ Plugging these into our arithmetic progression condition: $$2 \cdot \left(\frac{n(n - 1)}{2}\right) = n + \frac{n(n - 1)(n - 2)}{6}$$ Since $n > 3$ (and therefore $n \ne 0$), we can divide every term by $n$: $$n - 1 = 1 + \frac{(n - 1)(n - 2)}{6}$$ Multiply the entire equation by 6 to clear the fraction: $$6(n - 1) = 6 + (n^2 - 3n + 2) \implies 6n - 6 = n^2 - 3n + 8$$ Rearranging terms into standard quadratic form: $$n^2 - 9n + 14 = 0$$ Factoring this quadratic: $$(n - 2)(n - 7) = 0$$ The roots are $n = 2$ and $n = 7$. The problem specifies $n > 3$, so the only solution is $n = 7$!

Let's check $n = 7$: $$\binom{7}{1} = 7, \qquad \binom{7}{2} = 21, \qquad \binom{7}{3} = 35$$ The consecutive differences are $21 - 7 = 14$ and $35 - 21 = 14$!

In this lesson, we use CodeByMath's built-in comb(n, k) function to test progression conditions with loops, and search for larger combinations in arithmetic progression!