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Let the numbers be (x – 1), x and (x + 1) => (x – 1)2 + x2 + (x + 1)2 = 770 => x2 – 2x + 1 + x2 + x2 + 2x + 1 = 770 => 3x2 = 770 - 2 => x2 = 768/3 => x2 = 256 => x = 16 Required sum = (16 – 1) + (16 + 1) = 32
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