$4^{t h} t e r m$ of the geometric progression $x, x^{2} + 2, x^{3} + 10$ is

0

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6

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54

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27

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Solution

The correct option is C $54$
Given GP is,
$x, x^{2} + 22, x^{3} + 10$

where first term $a = x$ and common ration $d = \frac{x^{2} + 2}{x}$

Also
$\frac{x^{2} + 2}{x} = \frac{x^{3} + 10}{x^{2} + 2} ⟹ (x^{2} + 2)^{2} = x (x^{3} + 10)$

$⟹ x^{4} + 4 + 4 x^{2} = x^{4} + 10 x ⟹ 4 x^{2} - 10 x + 4 = 0$

$⟹ 4 x^{2} - 8 x - 2 x + 4 = 0 ⟹ (x - 2) (4 x - 2) = 0$

so we get
$x = 2 o r \frac{1}{2}$

Let we consider $x = 2$

Fourth term $= a r^{4 - 1} = a r^{3} = x \times (\frac{x^{2} + 2}{x})^{3}$
$= 2 \times (\frac{6}{2})^{3} = 2 \times 3^{3} = 2 \times 27 = 54$

hence we get forth term equal to $54$

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