[tex]a_n=a_1*r^{(n-1)}
\\\\
where \\\\
a_n=general \ term
\\a_1=first \ term
\\r=common\ ratio \\ (n-1)=same \ general \ term -1[/tex]
the data provided by the problem:
[tex]a_2=24 \\ a_5=-648
\\
\\ a_2
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_5
\\24=a_1*r^{(2-1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -648= a_1*r^{(5-1)}
\\24=a_1*r^1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -648 = a_1*r^4
\\\\a_1= \frac{24}{r} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_1= \frac{-648}{r^4} \\ \\ -----------------------
[/tex]
[tex]a_1=a_1 \\\\\frac{24}{r} = \frac{-648}{r^4}
\\\\ r^4(24)=r(-648)
\\24r^4 = -648r
\\24r^4+648r=0 \\
24r(r^3+27)=0
\\ 24r \left(r+3\right)\left(r^2-3r+9\right)=0 \\ \\
24r = 0 \ \ \ \ \ \ \ \ \ \ (r + 3) = 0 \ \ \ \ \ \ \ \ \ (r^2-3x+9)=0
\\\\ ----------------------------[/tex]
[tex]*24r=0
\\r= \frac{0}{24}
\\r=0
\\\\ *r+3=0
\\ r = 0-3
\\r=-3
\\\\ *r^2-3x+9=0 \\
\\x_{1,2}= \frac{-b+ \sqrt{b^2-4ac}}{2a}
\\\\x_{1}= \frac{-(-3)+\sqrt{(-3)^2-4(1)(9)}}{2(1)}
\\x_1=\frac{3+\sqrt{9-36}}{2}\\
\\x_1=\frac{3+\sqrt{-27}}{2}
\\x_1=\frac{3+3\sqrt{3}i}{2}
\\\\x_2=\frac{-(-3)-\sqrt{(-3)^2-4(1)(9)}}{2(1)}
\\ x_2=\frac{3-\sqrt{9-36}}{2}
\\x_2=\frac{3-\sqrt{-27}}{2} \\
x_2=\frac{3-3\sqrt{3i}}{2}[/tex]
---------------------------------------------------------
r ≠ 0
r = -3
[tex]a_n=a_1*r^{(n-1)} \\ \\
a_2 \\24=a_1*(-3)^{(2-1)}
\\24 = a_1*(-3)^1
\\24=a_1*-3
\\ \\ a_1= \frac{24}{-3}
\\a_1=-8
[/tex]
-------------------------------
[tex]a_{12}=a_1*r^{n-1}
\\a_{12}=-8*-3^{12-1}
\\a_{12}=-8*-3^{11}
\\a_{12}=-8*-177147
\\a_{12}=1417176
[/tex]
The 12th term is 1417176
