1. Stating the problem: We have a geometric progression (GP) where the third term is $\frac{9}{2}$ and the fifth term is $\frac{81}{8}$. We need to find: A. The common ratio $r$ B. The first term $a$ 2. Recall the formula for the $n^{th}$ term of a GP: $$a_n = ar^{n-1}$$ 3. Using the given information: - The third term: $$a_3 = ar^{2} = \frac{9}{2}$$ - The fifth term: $$a_5 = ar^{4} = \frac{81}{8}$$ 4. Divide the fifth term by the third term to eliminate $a$: $$\frac{a_5}{a_3} = \frac{ar^{4}}{ar^{2}} = r^{2} = \frac{\frac{81}{8}}{\frac{9}{2}} = \frac{81}{8} \times \frac{2}{9} = \frac{162}{72} = \frac{9}{4}$$ 5. Solve for $r$: $$r^{2} = \frac{9}{4} \implies r = \pm \frac{3}{2}$$ 6. Assuming the common ratio is positive (usual case in problems unless stated), take $$r = \frac{3}{2}$$ 7. Substitute $r$ back into the expression for the third term to find $a$: $$ar^{2} = \frac{9}{2} \implies a \left(\frac{3}{2}\right)^{2} = \frac{9}{2} \implies a \times \frac{9}{4} = \frac{9}{2}$$ 8. Solve for $a$: $$a = \frac{9}{2} \times \frac{4}{9} = 2$$ Final answers: A. Common ratio $r = \frac{3}{2}$ B. First term $a = 2$