Math Help - Goemetric seqeunce

1. Goemetric seqeunce question

a,b and c are three consecutive numbers terms of a goemetric sequence and that verify
a+b+c=19
2a+b-c=5
calculate a,b and c.

2. Re: Goemetric seqeunce

$a + ar + ar^2 = 19$

$2a + ar - ar^2 = 5$

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$3a + 2ar = 24$

$a(3+2r) = 24$

$a = \frac{24}{3+2r}$

$\frac{24}{3+2r} + \frac{24r}{3+2r} + \frac{24r^2}{3+2r} = \frac{19(3+2r)}{3+2r}$

$24 + 24r + 24r^2 = 57 + 38r$

$24r^2 - 14r - 33 = 0$

$(12r + 11)(2r - 3) = 0$

$r = -\frac{11}{12}$ or r = $\frac{3}{2}$

two possible solutions for a, b, c ...

$\frac{144}{7} , -\frac{132}{7} , \frac{121}{7}$

$4, 6, 9$