Geometric Mean Of Geometric Sequence

Let the common ratio of the geometric sequence by $r$ . Therefore $r^2=\dfrac{8}{18}=\dfrac{2}{3}$ .

The geometric mean of the sequence is the middle term which is $a_{50}$ .

Therefore $a_{50}=a_{49} \times r=18 \times \dfrac{2}{3}=\boxed{12}$ .

• Proof that the middle term of a geometric sequence is the middle term:

  Consider the geometric sequence $a,ar,ar^2,\ldots,ar^n$ where $a$ is the first term and $r$ is the common ratio.

  The geometric mean in defined by $\sqrt[n]{a \times ar \times ar^2 \times \ldots \times ar^n}=\sqrt[n] {a^nr^{(1+2+\ldots+n)}}=\sqrt[n]{a^nr^{\frac{n(n+1)}{2}}}=ar^{\frac{(n+1)}{2}}$ which is the middle term.