The next term of the geometric progression $x, x^{2} + 2, x^{3} + 10$ is _______

0

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6

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\frac{729}{16}

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54

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Solution

The correct option is D $54$
$x, x^{2} + 2, x^{3} + 3....$
If in $G P$
${(x^{2} + 2)}^{2} = (x^{3} + 10) x x^{4} + 4 + {4 x}^{2} = x^{4} + 10 x {4 x}^{2} - 10 x + 4 = 0 2 x^{2} - 5 x + 2 = 0 2 x^{2} - 4 x - x + 2 = 0 2 x (x - 2) - 1 (x - 2) = 0 (2 x - 1) (x - 2) = 0 x = \frac{1}{2}, 2 2, 6, 18....... a = 2 r = 3$
$∴$ Next term is $18 \times 3 = 54$

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