5 - Solving:
Answer: Formula P.G [tex]\boxed{A(n) = a*r^{n-1}}[/tex]
6- Solving
Collecting the data of this P.G., comes:
The 1st term [tex]a_{1} = 9[/tex]
The 2nd term [tex]a_{2} = 12[/tex]
The 3rd term [tex]a_{3} = 16[/tex]
The 4th term [tex]a_{4} = 21\frac{1}{3} = \frac{64}{3} [/tex]
It is a geometric progression (P.G), since the ratio of [tex] \frac{ a_{2} }{ a_{1} } = \frac{ a_{4} }{ a_{3}} [/tex]
The ratio [tex]r = \frac {a2}{a1}[/tex]
The ratio [tex]r = \frac {a4}{a3}[/tex]
The value of ratio, we have:
[tex]r = \frac {a2}{a1}[/tex]
[tex]r = \frac {12}{9}[/tex]
simplify by 3
[tex]r = \frac {12}{9}\frac{\div3}{\div3} \to\: \boxed{\boxed{r = \frac{4}{3}}}\end{array}}\qquad\quad\checkmark[/tex]
The value of ratio, we have:
[tex]r = \frac {a4}{a3}[/tex]
[tex]r = \frac{ \frac{64}{3} }{16} [/tex]
[tex]r = \frac{64}{3} * \frac{1}{16} = \frac{64}{48}[/tex]
Simplify by 16
[tex]r = \frac{64}{48} \frac{\div16}{\div16} \to\:\boxed{\boxed{r = \frac{4}{3} }}\end{array}}\qquad\quad\checkmark[/tex]
Answer:
[tex]\underline{Geometric\:with\:a\:common\:ratio\:of}[/tex] [tex]\boxed{\boxed{r = \frac{4}{3}}}[/tex]
