Question

If 1/x,1/8,4/x are in GP,then the positive value of x is:
A)1/16
B)16
C)61
D)32​

Answer 1

Answer:

16

Step-by-step explanation:

For three terms in GP ,

the square of middle term is equal to product of the two terms

So,

$\frac{1}{x} \times \frac{4}{x} = ({ \frac{1}{8} )}^{2} \\ \frac{4}{ {x}^{2} } = \frac{1}{64} \\ \frac{1}{ {x}^{2} } = \frac{1}{256} \\ \frac{1}{x} = \frac{1}{16} \\ x = 16$

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Answer 2

The required value of x "option B) 16" is correct.

Step-by-step explanation:

The given GP are:

$\dfrac{1}{x} ,\dfrac{1}{8},\dfrac{4}{x}$

Here, first term ($a_{1}$) = $\dfrac{1}{x}$, second term ($a_{2}$) = $\dfrac{1}{8}$ and

second term ($a_{3}$) = $\dfrac{4}{x}$

To find, the value of x = ?

We know that,

Common ratio (r) = $\dfrac{a_{2}}{a_{1}} =\dfrac{a_{3}}{a_{2}}$

∴ $\dfrac{\dfrac{1}{8}}{\dfrac{1}{x}} = \dfrac{\dfrac{4}{x}}{\dfrac{1}{8}}$

⇒ $\dfrac{x}{8} =\dfrac{4\times 8}{x}$

⇒ $\dfrac{x}{8} =\dfrac{32}{x}$

⇒ $x^2$ = 32 × 8 = 256

⇒ $x^2$ = $16^{2}$

⇒ x = 16

Thus, the required value of x "option B) 16" is correct.