Find the $\rm 2^{nd}$ term of a geometric sequence whose first and third terms are $\rm \frac{2}{3}$ and$\rm \frac{3}{2}$ respectively.
Solution
If the two numbers are a and b, then the geometric mean between them is given by $\rm GM = \sqrt{ab}$. If a and b are the first and the third terms of a Geometric Sequence, then the GM gives the second term of that sequence.
Given,
$\rm a = \frac{2}{3}$ and $\rm b = \frac{3}{2}$
By substituting the values into the formula, we get,
$\rm or, GM = \sqrt{\frac{2}{3} \cdot \frac{3}{2}}$
$\rm or, GM = \sqrt{ \frac{2}{3} \cdot \frac{3}{2} }$
$\rm or, GM = \sqrt{ 1 }$
$\rm or, GM = \sqrt{ \left ( 1\right )^{2} }$
$\rm \therefore GM = 1$
Hence, the required second term of the given Geometric Sequence is 1.