Respuesta :
Answer:
[tex]Tn\text{ = -}\frac{4}{3}(-3)^{n-1}[/tex]Explanation:
The nth term of a geometric sequence is expressed as;
[tex]\text{T}_{n\text{ }}=ar^{n-1}[/tex]If the second term is 4;
[tex]\begin{gathered} T_2=\text{ ar} \\ T_{2\text{ }}=\text{ ar= 4} \end{gathered}[/tex]If the third term is -12, hence;
[tex]\text{T}_{3\text{ }}=ar^2\text{= -12}[/tex]Solve equation 1 and 2 simultaneously for a and r
Divide both expressions
[tex]\begin{gathered} \frac{ar}{ar^2}=\text{ }\frac{4}{-12} \\ \frac{1}{r}=\frac{4}{-12} \\ r\text{ = -12/4} \\ r\text{ = -3} \end{gathered}[/tex]Get the first term;
Substitute r = -3 into 1;
Frm 1;
[tex]\begin{gathered} ar\text{ = 4} \\ -3a\text{ = 4} \\ a\text{ = }\frac{-4}{3} \end{gathered}[/tex]Get the explicit expression;
[tex]\begin{gathered} Tn=ar^{n-1} \\ Tn\text{ =-}\frac{4}{3}(-3)^{n-1^{}^{}} \end{gathered}[/tex]This gives the required answer
