1. **Problem (i):** A geometric sequence has first term $4$ and common ratio $6$. Find the minimum $n$ such that the $n^{th}$ term is greater than $10^{100}$. 2. The $n^{th}$ term of a geometric sequence is given by the formula: $$a_n = ar^{n-1}$$ where $a$ is the first term and $r$ is the common ratio. 3. Substitute $a=4$ and $r=6$: $$4 \times 6^{n-1} > 10^{100}$$ 4. Divide both sides by 4: $$\cancel{4} \times 6^{n-1} > \frac{10^{100}}{\cancel{4}}$$ $$6^{n-1} > 2.5 \times 10^{99}$$ 5. Take the logarithm base 6 of both sides: $$n-1 > \log_6(2.5 \times 10^{99})$$ 6. Using log properties: $$n-1 > \log_6(2.5) + \log_6(10^{99}) = \log_6(2.5) + 99 \log_6(10)$$ 7. Approximate values: $$\log_6(2.5) \approx 0.64, \quad \log_6(10) \approx 1.29$$ 8. Calculate: $$n-1 > 0.64 + 99 \times 1.29 = 0.64 + 127.71 = 128.35$$ 9. Therefore: $$n > 129.35$$ 10. Minimum integer $n$ is: $$\boxed{130}$$ --- 11. **Problem (ii):** A geometric sequence with first term $a$ and common ratio $r$ satisfies: - Second term is $-6$ - Sum to infinity is $25$ 12. The second term is: $$ar = -6$$ 13. The sum to infinity for $|r|<1$ is: $$S_\infty = \frac{a}{1-r} = 25$$ 14. From sum to infinity: $$a = 25(1-r)$$ 15. Substitute $a$ into second term equation: $$25(1-r)r = -6$$ 16. Expand: $$25r - 25r^2 = -6$$ 17. Rearrange to standard quadratic form: $$25r^2 - 25r - 6 = 0$$ 18. **(a) Shown:** $$25r^2 - 25r - 6 = 0$$ 19. **(b) Solve quadratic:** Use quadratic formula: $$r = \frac{25 \pm \sqrt{(-25)^2 - 4 \times 25 \times (-6)}}{2 \times 25} = \frac{25 \pm \sqrt{625 + 600}}{50} = \frac{25 \pm \sqrt{1225}}{50}$$ 20. Calculate: $$\sqrt{1225} = 35$$ 21. Solutions: $$r = \frac{25 + 35}{50} = \frac{60}{50} = 1.2$$ $$r = \frac{25 - 35}{50} = \frac{-10}{50} = -0.2$$ 22. **(c) Choose $r$:** Sum to infinity exists only if $|r| < 1$, so $r = -0.2$. 23. **(d) Find sum of first 4 terms:** First term: $$a = 25(1 - (-0.2)) = 25(1 + 0.2) = 25 \times 1.2 = 30$$ 24. Sum of first $n$ terms: $$S_n = a \frac{1 - r^n}{1 - r}$$ 25. Calculate $S_4$: $$S_4 = 30 \times \frac{1 - (-0.2)^4}{1 - (-0.2)} = 30 \times \frac{1 - 0.0016}{1 + 0.2} = 30 \times \frac{0.9984}{1.2} = 30 \times 0.832 = 24.96$$ 26. Final answer: $$\boxed{24.96}$$