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Algebra Difficulty 3.3 AMC 10/12 Find the answer

If x,2x+2,3x+3,x, 2x+2, 3x+3, \dots are in geometric progression, the fourth term is:

Pick one

Solution

Since we know the sequence is a geometric sequence, the ratio of consecutive terms is always the same number. Thus, we can set up an equation:
2x+2x=3x+32x+2\frac{2x+2}{x} = \frac{3x+3}{2x+2}.
Solving it, we get:
2x+2x=3x+32x+2\frac{2x+2}{x} = \frac{3x+3}{2x+2}
(2x+2)(2x+2)=(3x+3)(x)(2x+2)(2x+2) = (3x+3)(x)
4x2+8x+4=3x2+3x4x^2+8x+4 = 3x^2+3x
x2+5x+4=0x^2+5x+4 = 0
(x+4)(x+1)=0(x + 4)(x+1) = 0
x=4x = -4 or x=1x = -1
If x=1x=-1, the sequence has a 00 as the second term, which is not allowed in a geometric sequence, so it is an extraneous solution that came about because we cross-multiplied by (2x+2)(2x + 2), which is 00.
If x=4x=-4, we plug into x,2x+2,3x+3x, 2x +2, 3x + 3 to find the sequence starts as 4,6,9-4, -6, -9. The common ratio is 64=32\frac{-6}{-4} = \frac{3}{2}. The next term is 329=272=1312\frac{3}{2} \cdot -9 = \frac{-27}{2} = -13\frac{1}{2}, which is option (B)\textbf{(B)}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.