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