1964 AHSME Problems/Problem 6

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Since we know the sequence is a geometric sequence, we can set up an equation:

$\frac{2x+2}{x} = \frac{3x+3}{2x+2}$.

Solving it, we get:

$\frac{2x+2}{x} = \frac{3x+3}{2x+2}$


$(2x+2)(2x+2) = (3x+3)(x)$


$4x^2+8x+4 = 3x^2+3$


$x^2+8x+1 = 0$.

Applying the quadratic formula, we get

$\frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-8\pm \sqrt{8^2-4(1)(1)}}{2(1)} = \frac{-8\pm \sqrt{64-4}}{2} = \frac{-8\pm 2\sqrt{15}}{2} = -4\pm \sqrt{15}$.

From there, you can plug $x$ in to find out the answer.

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