Chapter 12.3, Problem 81AYU

To determine

To find: The value of $x$ .

Answer to Problem 81AYU

The value of x is $-4$ .

Explanation of Solution

Given information:

The given consecutive of geometric sequence are $x,x+2\text{ and }x+3$ .

Calculation:

Calculate the sequence,

Since, the ration must be the same of both sequences.

  $\frac{x+2}{x}=\frac{x+3}{x+2}$

Multiply both the sides of the equation by the LCM of the denominator, $\left(x+2\right)x$ .

  $\begin{array}{c}\left(\frac{x+2}{x}\right)x\left(x+2\right)=\left(\frac{x+3}{x+2}\right)x\left(x+2\right)\\ \left(x+2\right)\left(x+2\right)=\left(x+3\right)x\\ x^2+4x+4=x^2+3x\end{array}$

Subtract $x^2+3x-4$ from both sides,

  $\begin{array}{c}{x}^2+4x+4-x^2-3x-4=x^2+3x-x^2-3x-4\\ x=-4\end{array}$

Therefore, the value of $x$ is $-4$ .