A sequence is called an [i]arithmetic progression of the first order[/i] if the differences of the successive terms are constant. It is called an [i]arithmetic progression of the second order[/i] if the differences of the successive terms form an arithmetic progression of the first order. In general, for $k\geq 2$, a sequence is called an [i]arithmetic progression of the $k$-th order[/i] if the differences of the successive terms form an arithmetic progression of the $(k-1)$-th order.
The numbers
\[4,6,13,27,50,84\]
are the first six terms of an arithmetic progression of some order. What is its least possible order? Find a formula for the $n$-th term of this progression.
To determine the least possible order of the given sequence and find a formula for the \( n \)-th term, we need to analyze the differences between successive terms and identify the order of the arithmetic progression.

1. **Calculate the first differences:**
   \[
   \begin{aligned}
   a_2 - a_1 &= 6 - 4 = 2, \\
   a_3 - a_2 &= 13 - 6 = 7, \\
   a_4 - a_3 &= 27 - 13 = 14, \\
   a_5 - a_4 &= 50 - 27 = 23, \\
   a_6 - a_5 &= 84 - 50 = 34.
   \end{aligned}
   \]

2. **Calculate the second differences:**
   \[
   \begin{aligned}
   (a_3 - a_2) - (a_2 - a_1) &= 7 - 2 = 5, \\
   (a_4 - a_3) - (a_3 - a_2) &= 14 - 7 = 7, \\
   (a_5 - a_4) - (a_4 - a_3) &= 23 - 14 = 9, \\
   (a_6 - a_5) - (a_5 - a_4) &= 34 - 23 = 11.
   \end{aligned}
   \]

3. **Calculate the third differences:**
   \[
   \begin{aligned}
   [(a_4 - a_3) - (a_3 - a_2)] - [(a_3 - a_2) - (a_2 - a_1)] &= 7 - 5 = 2, \\
   [(a_5 - a_4) - (a_4 - a_3)] - [(a_4 - a_3) - (a_3 - a_2)] &= 9 - 7 = 2, \\
   [(a_6 - a_5) - (a_5 - a_4)] - [(a_5 - a_4) - (a_4 - a_3)] &= 11 - 9 = 2.
   \end{aligned}
   \]

Since the third differences are constant, the sequence is an arithmetic progression of the third order.

4. **Find the general formula for the \( n \)-th term:**
   Since the sequence is an arithmetic progression of the third order, the \( n \)-th term can be expressed as a cubic polynomial:
   \[
   a_n = An^3 + Bn^2 + Cn + D.
   \]

5. **Determine the coefficients \( A, B, C, \) and \( D \) using the given terms:**
   \[
   \begin{aligned}
   a_1 &= 4 \implies A(1)^3 + B(1)^2 + C(1) + D = 4, \\
   a_2 &= 6 \implies A(2)^3 + B(2)^2 + C(2) + D = 6, \\
   a_3 &= 13 \implies A(3)^3 + B(3)^2 + C(3) + D = 13, \\
   a_4 &= 27 \implies A(4)^3 + B(4)^2 + C(4) + D = 27.
   \end{aligned}
   \]

   This gives us the system of equations:
   \[
   \begin{aligned}
   A + B + C + D &= 4, \\
   8A + 4B + 2C + D &= 6, \\
   27A + 9B + 3C + D &= 13, \\
   64A + 16B + 4C + D &= 27.
   \end{aligned}
   \]

6. **Solve the system of equations:**
   Subtract the first equation from the second:
   \[
   7A + 3B + C = 2.
   \]

   Subtract the second equation from the third:
   \[
   19A + 5B + C = 7.
   \]

   Subtract the third equation from the fourth:
   \[
   37A + 7B + C = 14.
   \]

   Subtract the first new equation from the second new equation:
   \[
   12A + 2B = 5 \implies 6A + B = \frac{5}{2}.
   \]

   Subtract the second new equation from the third new equation:
   \[
   18A + 2B = 7 \implies 9A + B = \frac{7}{2}.
   \]

   Subtract the first simplified equation from the second simplified equation:
   \[
   3A = 1 \implies A = \frac{1}{3}.
   \]

   Substitute \( A = \frac{1}{3} \) back into \( 6A + B = \frac{5}{2} \):
   \[
   6 \left( \frac{1}{3} \right) + B = \frac{5}{2} \implies 2 + B = \frac{5}{2} \implies B = \frac{1}{2}.
   \]

   Substitute \( A = \frac{1}{3} \) and \( B = \frac{1}{2} \) back into \( 7A + 3B + C = 2 \):
   \[
   7 \left( \frac{1}{3} \right) + 3 \left( \frac{1}{2} \right) + C = 2 \implies \frac{7}{3} + \frac{3}{2} + C = 2.
   \]

   Convert to a common denominator:
   \[
   \frac{14}{6} + \frac{9}{6} + C = 2 \implies \frac{23}{6} + C = 2 \implies C = 2 - \frac{23}{6} = \frac{12}{6} - \frac{23}{6} = -\frac{11}{6}.
   \]

   Substitute \( A = \frac{1}{3} \), \( B = \frac{1}{2} \), and \( C = -\frac{11}{6} \) back into \( A + B + C + D = 4 \):
   \[
   \frac{1}{3} + \frac{1}{2} - \frac{11}{6} + D = 4.
   \]

   Convert to a common denominator:
   \[
   \frac{2}{6} + \frac{3}{6} - \frac{11}{6} + D = 4 \implies \frac{2 + 3 - 11}{6} + D = 4 \implies -1 + D = 4 \implies D = 5.
   \]

   Therefore, the formula for the \( n \)-th term is:
   \[
   a_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 - \frac{11}{6}n + 5.
   \]

   Simplifying, we get:
   \[
   a_n = \frac{1}{6}(2n^3 + 3n^2 - 11n + 30).
   \]

The final answer is \(\boxed{a_n = \frac{1}{6}(2n^3 + 3n^2 - 11n + 30)}\).