Question

The three numbers x-1,x and x+3 are consecutive terms of the G.P
A) find the value of x
B) find the common ratio

Answer 1

Answer:

The numeric value of used variable "x" is 3 / 2 .

Step-by-step explanation:

It is given that the three consecutive numbers of GP are x - 1 , x and x + 3.

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From the properties of geometric progressions, we know : -

$\dfrac{\text{First\:term}}{\text{Second\:erm}}=\text{Common Ratio}=\dfrac{\text{Second term}}{\text{Third term}}= \dfrac{\text{(n-1)th term}}{\text{nth term}}$

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Thus,

= >  ( x - 1 ) / x = x / ( x + 3 )

= > ( x - 1 ) ( x + 3 ) = x^2

= > x( x + 3 ) - 1( x + 3 ) = x^2

= >  x^2 + 3x - x - 3 = x^2

= > 2x - 3 =  0

= >  x = 3 / 2

Hence, numeric value of used variable "x" is 3 / 2 .

Answer 2

A) The value of x is $\dfrac{3}{2}$.

B) The common ratio is $\dfrac{1}{3}$.

Step-by-step explanation:

Given : The three numbers x-1, x and x+3 are consecutive terms of the G.P .

To find :

A) find the value of x

B) find the common ratio

Solution :

We know that the ratio between the two consecutive terms is same.

i.e. $\frac{a_1}{a_2}=\frac{a_2}{a_3}$

Here, $a_1=x-1,\ a_2=x,\ a_3=x+3$

$\frac{x-1}{x}=\frac{x}{x+3}$

$x^2=(x-1)(x+3)$

$x^2=x^2+3x-x-3$

$2x=3$

$x=\frac{3}{2}$

So, the value of x is $\dfrac{3}{2}$.

$x-1=\frac{3}{2}-1=\frac{1}{2}$

The common ratio is $r=\frac{a_1}{a_2}$

$r=\frac{x-1}{x}$

$r=\dfrac{\frac{1}{2}}{\frac{3}{2}}$

$r=\frac{1}{3}$

So, the common ratio is $\dfrac{1}{3}$.

#Learn more

If x,x+2,x+3, are consecutive terms of G.p find x.

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