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a, b and c are all different numbers in AP. If the roots of ax^2 + bx + c = 0 are p and q and 1/p + 1/q, p + q, p^2 + q^2 is in GP, then find a/c. So I started by simplifying the GP to p + q/pq, p + q, p^2 + q^2 => -b/c, -b/a, (b^2 - 2ac)/a^2 And then I imposed the condition that -b/a * c/a = (b^2 - 2ac)/a^2 => -bc = b^2 - 2ac => b^2 + bc - 2ac = 0 Now since a, b, c are in AP, I substituted b = (a+c)/2 => [(a+c)/2]^2 + c * [(a+c)/2] - 2ac = 0 => (a^2 + c^2 + 2ac)/4 + (ac + c^2)/2 - 2ac = 0 => a^2 + c^2 + 2ac + 2ac + 2c^2 - 8ac = 0 => a^2 + 3c^2 - 4ac = 0 => (a/c)^2 + 3 - 4(a/c) = 0 => (a/c)^2 - 3(a/c) - (a/c) + 3 = 0 => a/c = 1, 3 We eliminate 1.
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