Respuesta :
Using geometric sequence concepts, it is found that:
- a) The first three teams are 9, 6 and 4.
- b) It is the 8th term.
What is a geometric sequence?
- A geometric sequence is a sequence in which the result of the division of consecutive terms is always the same, called common ratio q.
The nth term of a geometric sequence is given by:
[tex]a_n = a_1q^{n-1}[/tex]
- In which [tex]a_1[/tex] is the first term.
In this problem, the sequence is: {2p - 1, p + 1, p - 1}.
Since it is geometric, the ratio of consecutive terms is the same, that is:
[tex]\frac{p - 1}{p + 1} = \frac{p + 1}{2p - 1}[/tex]
Applying cross multiplication:
[tex](2p - 1)(p - 1) = (p + 1)^2[/tex]
[tex]2p^2 - 3p + 1 = p^2 + 2p + 1[/tex]
[tex]p^2 - 5p = 0[/tex]
[tex]p(p - 5) = 0[/tex]
[tex]p \neq 0[/tex], hence:
[tex]p - 5 = 0[/tex]
[tex]p = 5[/tex]
Item a:
[tex]2p - 1 = 2(5) - 1 = 9[/tex]
[tex]p + 1 = 5 + 1 = 6[/tex]
[tex]p - 1 = 5 - 1 = 4[/tex]
Hence:
The first three teams are 9, 6 and 4.
Item b:
- The first term is [tex]a_1 = 9[/tex].
- The common ratio is [tex]q = \frac{6}{9} = \frac{2}{3}[/tex].
Then:
[tex]a_n = a_1q^{n-1}[/tex]
[tex]a_n = 9\left(\frac{2}{3}\right)^{n-1}[/tex]
Now we have to find n for which [tex]a_n = \frac{128}{243}[/tex]
[tex]a_n = 9\left(\frac{2}{3}\right)^{n-1}[/tex]
[tex]\frac{128}{243} = 9\left(\frac{2}{3}\right)^{n-1}[/tex]
[tex]\left(\frac{2}{3}\right)^{n-1} = \frac{128}{243 \times 9}[/tex]
[tex]\left(\frac{2}{3}\right)^{n-1} = \frac{2^7}{3^5 \times 3^2}[/tex]
[tex]\left(\frac{2}{3}\right)^{n-1} = \left(\frac{2}{3}\right)^{7}[/tex]
[tex]n - 1 = 7[/tex]
[tex]n = 8[/tex]
Hence:
It is the 8th term.
You can learn more about geometric sequence concepts at brainly.com/question/11847927
