The square of the sum is $ (4a)^2 = 16a^2 $.
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- Solution: Let the four terms be $ a - \frac{3d}{2},\ a - \frac{d}{2},\ a + \frac{d}{2},\ a + \frac{3d}{2} $. This ensures the terms are in arithmetic progression with common difference $ d $. The first term is $ a - \frac{3d}{2} $, and the last term is $ a + \frac{3d}{2} $. Their sum of squares is:
- \left(a - \frac{3d}{2}\right)^2 + \left(a + \frac{3d}{2}\right)^2 = 2a^2 + 2\left(\frac{9d^2}{4}\right) = 2a^2 + \frac{9d^2}{2}.
- The sum of all four terms is:
- 2a^2 + \frac{9d^2}{2} = 16a^2
- \Rightarrow \frac{9d^2}{2} = 14a^2
- \Rightarrow d^2 = \frac{28a^2}{9}.
