1,-2,4,-8,16 and 16,-8,4,-2,1
a=first term, q=common ratio, n=number of terms.
Q=qn.
q=1 is impossible, as in that case an=11, a2n=341, and hence a=31, n=11/31, which contradicts to the condition a3n=3641
Therefore, the following relationships are true:
(1)-> a{Q-1 / q-1}=11
(2)-> a^2{Q^2-1 / q^2-1}=341
(3)-> a^3{Q^3-1 /r q^3-1}=3641
Divide (2) and (3) by (1).
Then a{Q+1 / q+1}=31 a^2{Q^2+Q+1 / q^2+q+-1}=331.
Hence a(Q-1)=11(q-1) (Q+1)=31(q+1) thus, a=10q+21 and Q={21q+10 / 10q+21}
Substitute the result into the equality a2(Q2+Q+1)=331(q2+q+1).
possible values of q : the roots of the equation 2q2+5q+2=0: q=-2 and q=-1/2.
Hence either a=1, Q=-32, n=5, or a=16, Q=-1/32, n=5.
Thus the solutions of the problem are the following two sequences: 1,-2,4,-8,16 and 16,-8,4,-2,1