Respuesta :
Given:
[tex]\begin{gathered} a3=36 \\ \\ r=-\frac{3}{4} \end{gathered}[/tex]Required:
To find the rule for the nth term. Then graph the first six terms of the sequence.
Explanation:
The nth term of the geometric sequence is given by
[tex]a_n=ar^^{n-1}[/tex]Here
[tex]\begin{gathered} a_n=36 \\ n=3 \\ r=-\frac{3}{4} \end{gathered}[/tex]Now we have to find the value of a,
[tex]\begin{gathered} 36=a(-\frac{3}{4})^{3-1} \\ \\ 36=a(-\frac{3}{4})^2 \\ \\ 36=a(\frac{9}{16}) \\ \\ a=\frac{576}{9} \\ \\ a=64 \end{gathered}[/tex]The first term is 64, therefore the formula is
[tex]a_n=64(-\frac{3}{4})^{n-1}[/tex]Now the first six terms of the sequence are
[tex]\begin{gathered} a1=64(-\frac{3}{4})^0 \\ \\ a1=64 \\ \\ a2=64(-\frac{3}{4})^1 \\ \\ =64(-\frac{3}{4}) \\ \\ =-48 \\ \\ a3=64(-\frac{3}{4})^2 \\ \\ =64(\frac{9}{16}) \\ \\ =36 \\ \\ a4=64(-\frac{3}{4})^3 \\ \\ =64(-\frac{27}{64}) \\ \\ =-27 \\ \\ a5=64(-\frac{3}{4})^4 \\ \\ =64(\frac{81}{256}) \\ \\ =\frac{81}{4} \\ \\ a6=64(-\frac{3}{4})^5 \\ \\ =-\frac{243}{16} \end{gathered}[/tex]Therefore the first 6th terns are
[tex]64,-48,36,-27,\frac{81}{4},-\frac{243}{16}[/tex]Final Answer:
The rule for the nth term is
[tex]a_n=64(-\frac{3}{4})^{n-1}[/tex]The first 6th terns are :
[tex]64, -4,836, -27, \frac{81}{4}, -\frac{243}{16}[/tex]