Question

What is the 12th term in the geometric sequence in which a2 is 24 and A5 is -648

Answer 1

`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`

the data provided by the problem:


`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= (24)/(r) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_1= (-648)/(r^4) \\ \\ -----------------------`


`a_1=a_1 \\\\(24)/(r) = (-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 \\\\ ----------------------------`


`*24r=0 \\r= (0)/(24) \\r=0 \\\\ *r+3=0 \\ r = 0-3 \\r=-3 \\\\ *r^2-3x+9=0 \\ \\x_(1,2)= (-b+ √(b^2-4ac))/(2a) \\\\x_(1)= (-(-3)+√((-3)^2-4(1)(9)))/(2(1)) \\x_1=(3+√(9-36))/(2)\\ \\x_1=(3+√(-27))/(2) \\x_1=(3+3√(3)i)/(2) \\\\x_2=(-(-3)-√((-3)^2-4(1)(9)))/(2(1)) \\ x_2=(3-√(9-36))/(2) \\x_2=(3-√(-27))/(2) \\ x_2=(3-3√(3i))/(2)`

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r ≠ 0
r = -3


`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= (24)/(-3) \\a_1=-8`

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`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`

The 12th term is 1417176