sequence

[red and white kop!]

find the value of x for which the numbers x+1, X+3 and x+7 are in geometrical progression.
ok so i tried using (x+7/X+3)= (x+3/x+1)^2 but im really not sure that this is the right method! anyway i got the answer wrong. any help?

 

[Deleted member 4993]

find the value of x for which the numbers x+1, X+3 and x+7 are in geometrical progression.
ok so i tried using (x+7/X+3)= (x+3/x+1)^2 but im really not sure that this is the right method! anyway i got the answer wrong. any help?

\(\displaystyle x \, + \, \frac{7}{X} \, + \, 3 \, = \, \left (x \, + \, \frac{3}{x} \, + \, 1\right )^2\)

is incorrect relationship to use.

The correct equation to use would be

(x+3)/(x+1) = (x+7)/(x+3)

or

(x+3)^2 = (x+1)(x+7)

 

[soroban]

Hello, red and white kop!

Find the value of \(\displaystyle x\) for which the numbers \(\displaystyle x+1,\: x+3,\: x+7\) are in geometrical progression.

\(\displaystyle \text{Since they are a geometric progression, the common ratio is: }\;r \;=\;\frac{x+3}{x+1} \;=\;\frac{x+7}{x+3}\)

\(\displaystyle \text{Then we have: }\x+3)(x+3) \:=\x+1)(x+7) \quad\Rightarrow\quad x^2 + 6x + 9 \:=\:x^2 + 8x + 7 \quadf\Rightarrow\quad 2 \:=\:2x\)

\(\displaystyle \text{Therefore: }\;x \:=\:1\)

 

[red and white kop!]

thanks