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Home » Unlock the Secret to Perfect Instagram Highlight Covers – No Design Skills Needed! - Crosslake

Unlock the Secret to Perfect Instagram Highlight Covers – No Design Skills Needed! - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 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 all: $ 4a $, square: $ 16a^2 $. Set equal:

📌 2a^2 + 18d^2 = 16a^2 \Rightarrow 18d^2 = 14a^2 \Rightarrow \frac{d^2}{a^2} = \frac{14}{18} = \frac{7}{9}.

📌 Third term: $ a + d $, first: $ a - 3d $. Ratio:

📌 \frac{a + d}{a - 3d} = \frac{1 + \frac{d}{a}}{1 - 3\frac{d}{a}}.

📌 Let $ x = \frac{d}{a} $, $ x^2 = \frac{7}{9} \Rightarrow x = \pm \frac{\sqrt{7}}{3} $. Try $ x = \frac{\sqrt{7}}{3} $:

📌 \frac{1 + \frac{\sqrt{7}}{3}}{1 - \sqrt{7}} = \frac{\frac{3 + \sqrt{7}}{3}}{1 - \sqrt{7}} = \frac{3 + \sqrt{7}}{3(1 - \sqrt{7})}.

📌 Numerator: $ (3 + \sqrt{7})(1 + \sqrt{7}) = 3 + 3\sqrt{7} + \sqrt{7} + 7 = 10 + 4\sqrt{7} $.

📌 Denominator: $ 3(1 - 7) = 3(-6) = -18 $.

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