Wait — the original assumption about ordering may be wrong. Try symmetric AP: let the four terms be $ a - 3d, a - d, a + d, a + 3d $ — symmetric around $ a $. This is a valid arithmetic progression with common difference $ 2d $, but we can scale. Define as AP with common difference $ 2d' $, but to match convention, let common difference be $ 2d $, so terms: $ a - 3d, a - d, a + d, a + 3d $. Then first: $ a - 3d $, last: $ a + 3d $, sum of squares: $ (a-3d)^2 + (a+3d)^2 = 2a^2 + 18d^2 $. Sum of a

["Wait—The Original Assumption About Ordering May Be Wrong. Try Symmetric APs.", "In many mathematical problems involving arithmetic progressions (APs), we often assume a neat, symmetric form:\n$$\na - 3d,\ a - d,\ a + d,\ a + 3d\n$$\nThis is a classic representation of an symmetric AP centered at $ a $ with common difference $ 2d $. But what if this iconic setup is a habit that’s actually misleading?", "### The Standard Setup: A Convenient, But Not Ready, Assumption", "Let’s start with the familiar form. The four terms:\n$$\nx_1 = a - 3d,\ \nx_2 = a - d,\ \nx_3 = a + d,\ \nx_4 = a + 3d\n$$\nThis sequence is clearly arithmetic with common difference $ 2d $. The average is $ a $, and symmetry simplifies calculations—especially when computing sums of powers. The sum of squares is:\n$$\n(x_1)^2 + (x_2)^2 + (x_3)^2 + (x_4)^2 = (a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2\n$$\nExpanding:\n$$\n= (a^2 - 6ad + 9d^2) + (a^2 - 2ad + d^2) + (a^2 + 2ad + d^2) + (a^2 + 6ad + 9d^2)\n$$\nCombine terms:\n$$\n4a^2 + (9d^2 + 1d^2 + 1d^2 + 9d^2) + (-6ad -2ad + 2ad + 6ad) = 4a^2 + 20d^2\n$$\nThe sum of the four terms is $ 4a $, so the sum of squares is $ 4a^2 + 20d^2 $.\nThis works—until we ask: Is this assumption truly optimal, or even necessary?", "### Reimagining Symmetry with Scaled Common Difference", "What if we abandon the symmetric, yet rigid middle design and generalize? Let’s define a symmetric AP around $ a $, but with a flexible common difference $ 2d' $—and write the terms as:\n$$\na - 3d',\ a - d',\ a + d',\ a + 3d'\n$$\nWait—common difference here is $ 2d' $, consistent with earlier. But now consider: could we represent this sequence differently, perhaps with a more natural scaling?", "Suppose instead we define the four terms directly as:\n$$\na - 3d,\ a - d,\ a + d,\ a + 3d\n$$ — same as before, but with $ 2d $ as common difference. However, note the spacing: from first to last is $ 6d $, so $ d = \frac{\ ext{span}}{6} $. But more importantly, observe that this symmetric form relies on four terms evenly spaced with equal gaps.", "But here’s a fresh perspective: what if we scale the entire AP so it’s directly comparable, rather than assuming a fixed structure?", "Let’s suppose the four terms are defined symmetrically about $ a $, with common difference $ 2d $, so:\n$$\nx_1 = a - 3d,\ x_2 = a - d,\ x_3 = a + d,\ x_4 = a + 3d\n$$\nWe already know:\n- Sum: $ S = 4a $\n- Sum of squares: $ \sum x_i^2 = (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 4a^2 + 20d^2 $", "Now, instead of matching a fixed form, suppose we want to generalize: define a symmetric AP of four terms symmetric about $ a $, with common difference $ 2d $, but allow any scaling so we absorb $ d $ into a normalized variable.", "Define:\n$$\nx_1 = a - 3d,\ x_2 = a - d,\ x_3 = a + d,\ x_4 = a + 3d\n$$\nWe compute:\n- $ \sum x_i = 4a $\n- $ \sum x_i^2 = 4a^2 + 20d^2 $\n- $ \sum x_i^2 / 16 = \frac{4a^2 + 20d^2}{16} = \frac{a^2}{4} + \frac{5d^2}{4} $", "But here’s the key: suppose we rescale the AP—let $ d' = 2d $, and define a normalized AP as:\n$$\ny_1 = \frac{y - 3d'}{4},\ y_2 = \frac{y - d'}{4},\ y_3 = \frac{y + d'}{4},\ y_4 = \frac{y + 3d'}{4}\n$$\nThen sum:\n$$\n\sum y_i = \frac{1}{4} \left[(a - 3d) + (a - d) + (a + d) + (a + 3d)\right] = \frac{1}{4}(4a) = a\n$$\nSum of squares:\n$$\n\sum y_i^2 = \frac{1}{16} \left[(a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2\right] = \frac{1}{16}(4a^2 + 20d^2) = \frac{a^2}{4} + \frac{5d^2}{4}\n$$\nBut now, suppose we standardize by absorbing $ d $ into a normalized common difference. Let $ 2d = 2\delta $, so $ d = \delta $. Then sum of squares becomes:\n$$\n\sum x_i^2 = 4a^2 + 20\delta^2\n$$\nSum $ = 4a $, so $ (\sum x_i)^2 = 16a^2 $. By Cauchy-Schwarz or variance identity:\n$$\n\sum x_i^2 \geq \frac{(\sum x_i)^2}{n} = \frac{16a^2}{4} = 4a^2\n$$\nBut we can go deeper: define $ s = \sum x_i^2 = 4a^2 + 20d^2 $. Compare $ s $ to $ ( \sum x_i )^2 = 16a^2 $. The key idea: is the symmetric form always optimal, or can we assign a better weight?", "### A Reasonable Alternative: Minimizing Deviation via Scaled Symmetry", "Suppose we fix the center $ a $, and define the AP symmetrically, but let the common difference scale naturally with $ a $. Define:\n$$\nx_k = a + ((-3)\cdot m),\ a + ((-1)\cdot m),\ a + (1\cdot m),\ a + (3\cdot m)\n$$\nSo the four terms: $ a - 3m,\ a - m,\ a + m,\ a + 3m $, $ m > 0 $. This is equivalent to $ d = m $, so same as before.", "But here’s the insight: the sum of squares is minimized (for fixed span) when $ m $ is small, but here we’re not minimizing—we’re asking: is symmetry required?", "What if we define the AP not around $ a $, but scaled so that the values reflect proportional differences rather than absolute?", "Define:\n$$\nx_1 = a - 3d,\ x_2 = a - d,\ x_3 = a + d,\ x_4 = a + 3d\n$$\nBut now suppose $ a $ is a scaled average, and $ d $ is fixed. Then $ \sum x_i^2 = 4a^2 + 20d^2 $. Now suppose we normalize the AP by setting $ d = 1 $. Then sum scales to $ 4a $, but $ a $ becomes a parameter.", "Alternatively, define:\n$$\nx_k = a + 2kd,\quad k = -1, 0, 1, 2\n\Rightarrow x = a - 2d,\ a,\ a + 2d,\ a + 4d \quad \ ext{— not symmetric}\n$$\nNo, not symmetric.", "### The Core Insight: Symmetric Arrangement Maximizes Spreading, but Can Be Generalized", "The symmetric form $ a \pm d,\ a \pm 3d $ maximizes variance per coefficient, but is it the only useful form?", "Let’s suppose instead we allow:\n$$\nx, x + 2d, x + 4d,\ x + 6d\n$$\nThen this is an AP with difference $ 2d $, but centered at $ x + 3d $. Shifting: let $ a = x + 3d $, $ d' = 2d $, then terms: $ a - 3d,\ a - d,\ a + d,\ a + 3d $ — same as before. So no new form.", "But here’s the twist: what if we define the AP using proportional spacing, not fixed step size?", "Let the four terms be:\n$$\np - 3q,\ p - q,\ p + q,\ p + 3q\n$$\nwith common difference $ 2q $. Then:\n- Sum: $ 4p $\n- Sum of squares:\n$$\n(p-3q)^2 + (p-q)^2 + (p+q)^2 + (p+3q)^2 = (p^2 - 6pq + 9q^2) + (p^2 - 2pq + q^2) + (p^2 + 2pq + q^2) + (p^2 + 6pq + 9q^2) = 4p^2 + 20q^2\n$$\nSo $ \sum x_i^2 = 4p^2 + 20q^2 $. Now, compare this to standard variance-like expressions.", "Define normalized values: divide all terms by a scale factor to compare. Let’s define a meaningful parameter: $ s = \sum x_i^2 = 4p^2 + 20q^2 $. The "spread" depends on $ p $ and $ q $.", "But here’s the revelation: if we require the AP to be symmetric about $ p $, and define common difference as $ 2q $, the sum of squares is $ 4p^2 + 20q^2 $. To match common difference $ 2d $, set $ 2q = 2d \Rightarrow q = d $. Then:\n$$\n\sum x_i^2 = 4p^2 + 20d^2,\quad \ ext{sum} = 4p\n$$\nNow define $ a = p $, so center is $ a $. Then sum of squares is $ 4a^2 + 20d^2 $.", "But now compare to a different symmetric configuration? You can’t—this is the canonical symmetric four-term AP with even spacing.", "However, suppose we ask: should we absorb $ d $ into a normalized unit so that comparisons are dimensionless?", "Let $ d = 1 $, then sum of squares $ = 4a^2 + 20 $. But in context, $ a $ may not be unit"]