What is the first term of the geometric sequence presented in the table below?

12    =  a₁(r)²   →    a₁  = 12/r²      (1)
-324 =  a₁(r)⁵     (2)

Putting  (1)  into (2), we have

-324  = (12/r²)(r)⁵

-324/12 = r³

-27  = r³    

r = (-27)^1/3 = -3

So

a₁  = 12/(-3)²  =  12/9  =  4/3

What is the first term of the geometric sequence presented in the table below?
 

n      3      6

a_n  12   −324

$\small{ \boxed{~ \text{geometric sequence: } \quad a_k = a_i^{ \frac{j-k}{j-i} }\cdot a_j^{ \frac{k-i}{j-i} } ~} }$

$\begin{array}{rcll} a_3 &=& a_i = 12 \qquad & i = 3\\ a_6 &=& a_j =-324 \qquad & j = 6\\ a_1 &=& a_k = \ ? \qquad & k = 1 \end{array}$

$\begin{array}{rcll} a_1 &=& 12 ^{ \left( \frac{6-1}{6-3} \right)} \cdot (-324)^{ \left( \frac{1-3}{6-3} \right) } \\ a_1 &=& 12 ^{\left( \frac{5}{3} \right) } \cdot (-324)^{ \left( \frac{-2}{3} \right) } \\ a_1 &=& \frac{ 12 ^{\left( \frac{5}{3} \right) } } { (-324)^{ \left( \frac{2}{3} \right) } }\\ a_1 &=& \sqrt[3]{ \frac{ 12^5 } { (-324)^{2} } } \\ a_1 &=& \sqrt[3]{ \frac{ 248832 } { 104976 } } \\ a_1 &=& \sqrt[3]{ 2.37037037037 } \\ a_1 &=& 1.33333333333 \\ a_1 &=& \frac43 \\ \end{array}$