Sum of the first ten terms of a Geometric Progression

a₁ = 2x, a₂ = 4x + 14, a₃ = 20x - 14

For GP
$\dfrac{a_2}{a_1} = \dfrac{a_3}{a_2}$

$\dfrac{4x + 14}{2x} = \dfrac{20x - 14}{4x + 14}$

$(4x + 14)^2 = 2x(20x - 14)$

$16x^2 + 112x + 196 = 40x^2 - 28x$

$24x^2 - 140x - 196 = 0$

$6x^2 - 35x - 49 = 0$

$(x - 7)\left(x + \dfrac{7}{6}\right) = 0$

$x = 7 ~ \text{and} ~ -\dfrac{7}{6}$
 

The sum of first n terms of GP is...
$S_n = \dfrac{a_1(r^n - 1)}{r - 1}$
 

For x = 7

$a_1 = 2(7) = 14$

$a_2 = 4(7) + 14 = 42$

$r = \dfrac{42}{14} = 3$
 

$S_{10} = \dfrac{14(3^{10} - 1)}{3 - 1}$

$S_{10} = 413,336$   ←   not in the choices

 

For x = -7/6

$a_1 = 2(-7/6) = -7/3$

$a_2 = 4(-7/6) + 14 = 28/3$

$r = \dfrac{28/3}{-7/3} = -4$
 

$S_{10} = \dfrac{-\frac{7}{3}[ \, (-4)^{10} - 1 \, ]}{-4 - 1}$

$S_{10} = 489,335$   ←   answer