1. **Problem Statement:** Given a geometric progression (GP) with the first term $a_1 = 108$ and the fifth term $a_5 = \frac{4}{3}$, find: (i) The two possible values of the common ratio $r$. (ii) The corresponding values of the 2nd, 3rd, and 4th terms. 2. **Formula and Rules:** In a GP, the $n$th term is given by: $$a_n = a_1 \times r^{n-1}$$ where $a_1$ is the first term and $r$ is the common ratio. 3. **Find the common ratio $r$:** Given $a_5 = \frac{4}{3}$, substitute $n=5$: $$\frac{4}{3} = 108 \times r^{4}$$ Divide both sides by 108: $$r^{4} = \frac{4}{3} \div 108 = \frac{4}{3} \times \frac{1}{108} = \frac{4}{324} = \frac{1}{81}$$ 4. **Solve for $r$:** $$r^{4} = \frac{1}{81}$$ Since $81 = 3^4$, we have: $$r^{4} = \left(\frac{1}{3}\right)^4$$ Taking the fourth root: $$r = \pm \frac{1}{3}$$ 5. **Find the unknown terms for each $r$:** - For $r = \frac{1}{3}$: $$a_2 = 108 \times \frac{1}{3} = 36$$ $$a_3 = 108 \times \left(\frac{1}{3}\right)^2 = 108 \times \frac{1}{9} = 12$$ $$a_4 = 108 \times \left(\frac{1}{3}\right)^3 = 108 \times \frac{1}{27} = 4$$ - For $r = -\frac{1}{3}$: $$a_2 = 108 \times \left(-\frac{1}{3}\right) = -36$$ $$a_3 = 108 \times \left(-\frac{1}{3}\right)^2 = 108 \times \frac{1}{9} = 12$$ $$a_4 = 108 \times \left(-\frac{1}{3}\right)^3 = 108 \times \left(-\frac{1}{27}\right) = -4$$ **Final answers:** - Common ratios: $r = \frac{1}{3}$ or $r = -\frac{1}{3}$ - Corresponding terms: - For $r=\frac{1}{3}$: $36, 12, 4$ - For $r=-\frac{1}{3}$: $-36, 12, -4$