We have two geometric sequences of positive real numbers:
6,a,b and a,b,1296
Solve for a
geometric sequence is \(a_n=a_1*q^{n-1}\), and so fore \(a_1=a_1*q^{1-1}, a_2=a_1*q^{2-1}, a_3=a_1*q^{3-1}....\).
now we know that if there is a geometric sequence \(6*q^1=a, 6*q^2=b, 6*q^3=1296\), and a sequence of \(a_1*q^1=a, a_1*q^2=b, a_1*q^3=1296 \), we can group them together and form a one sequnce \(6*q^0=6, 6*q^1=a, 6*q^2=b, 6*q^3=1296\),. Lets solve then q out of the final form \(6*q^3=1296, 1296/6=q^3, 216^{1/3}=q, q=6\). Now that we know that a is \(a_1*q^1=a, a_1=6, q=6, 6*6^1=36\)
