Respuesta :

Answer:

x = ± [tex]\sqrt{2}[/tex]

Step-by-step explanation:

given the 3 terms of a geometric sequence

[tex]\sqrt{x-1}[/tex], 1, [tex]\sqrt{x+1}[/tex], then the common ratio r

r = [tex]\frac{\sqrt{(x+1)} }{1}[/tex] = [tex]\frac{1}{\sqrt{(x-1)} }[/tex] ( cross- multiply )

[tex]\sqrt{(x+1)(x-1)}[/tex] = 1 ( square both sides )

(x + 1 )(x - 1 ) = 1 ← expand factors on left side )

x² - 1 = 1 ( add 1 to both sides )

x² = 2 ( take the square root of both sides )

x = ± [tex]\sqrt{2}[/tex]

The value of x is ±[tex]\sqrt{2}[/tex].

What is a Geometric Sequence?

  • It is a sequence, where the ratio of two consecutive terms in a sequence are constant.
  • It is a finite or infinite sequence.
  • Geometric Sequence G is given by: G = a, ar², ar³,..... (a = first term; r = constant ratio)

Given:

Geometric Sequence with first three terms.

[tex]\sqrt{x-1},1, \sqrt{x+1[/tex]

We have to find the value of x.

Now, if the given sequence is geometric then we can find the common ratio of the sequence.

The common ratio is given by:

r = [tex]\frac{1}{\sqrt{x-1} }[/tex] and r = [tex]\frac{\sqrt{x+1} }{1}[/tex]

Now, we can equate the both the values of r to solve for the value of x.

[tex]\frac{\sqrt{x+1} }{1} = \frac{1}{\sqrt{x-1} }[/tex]

Now, multiplying the numerator on both sides by [tex]\sqrt{x-1}[/tex], we get:

[tex]\sqrt{x+1} \sqrt{x-1} =1[/tex]

[tex]\sqrt{(x-1)(x+1)}=1[/tex]

[tex]\sqrt{x^2-1} =1[/tex]

Now, squaring on both sides of the equation, we get:

⇒ x² - 1 = 1

⇒ x² = 2

Now, taking square root on both the sides, we get:

⇒ x = ±[tex]\sqrt{2}[/tex]

Therefore, for a given geometric sequence the value of x will be ±[tex]\sqrt{2}[/tex].

Learn more about the Geometric Sequence here: https://brainly.com/question/14565307

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