Question

For what values of x is -2/7 , x , -7/2 are in geometric progression

Answer 1

Answer:

For a sequence to be in Geometric Progression, the common ratio (r) must be equal.

Let a, b, c be in a Geometric Progression.

Then, the condition to be satisfied is:

$\implies \boxed{ \dfrac{b}{a} = \dfrac{c}{b} = r}$

where, 'r' is the common ratio.

Now, in the given question, the sequence is:

⇒ -2/7, x, -7/2

Applying the condition of G.P we get:

$\implies \dfrac{ x }{ -2/7} = \dfrac{-7/2}{x}\\\\\\\implies x^2 = \dfrac{-7}{2} \times \dfrac{-2}{7}\\\\\\\implies x^2 = 1\\\\\\\implies \boxed{x = \sqrt{1} = \pm 1}$

Hence the possible values of 'x' for the given sequence to be in Geometric Progression is: +1 & -1.

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

For what values of x is -2/7 , x , -7/2 are in geometric progression.

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$\bf \:\huge \red{AηsωeR } ✍$

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Concept used :-

If a, b c are in geometric progression or G. P. then

$\sf \:  \dfrac{b}{a} \: = \dfrac{c}{b}$

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$\large\underline\purple{\bold{Solution :- }}$

$\bf \: Since, \: \dfrac{2}{7} \: , \: x \: , \: - \: \dfrac{7}{2} \: are \: in \: geometric \: progression.$

$\bf \:  ⟼ \therefore \: \dfrac{x}{\dfrac{ - 2}{7} } \: = \: \dfrac{ - 7}{\dfrac{2}{x} }$

$\bf \:  ⟼ \dfrac{ \cancel{ - 7}x}{ \cancel{2}} = \dfrac{ \cancel{ - 7}}{ \cancel{2}x}$

$\bf\implies \: {x}^{2} = 1$

$\bf\implies \:x \: = \: \pm \: 1$

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