Arithmetic and geometric progressions: EJC 2024 Promo Q7

(ai)

If $r = 1$, all the terms of the sequence will be equal to the first term $a$. Then the sum of the first 10 terms will be $10a$, which is only 2 times the sum of the first 5 terms $5a$.

(aii)

Using sum of G.P. formula, $$\begin{aligned} a\times\frac{1 - r^{10}}{1 - r} &= 33\times a\times\frac{1 - r^5}{1 - r}\\ 1 - r^{10} &= 33 - 33r^5\\ \left(1 - r^5\right)\left(1 + r^5\right) &= 33\left(1 - r^5\right)\\ \left(1 - r^5\right)\left(1+r^5-33\right) &= 0\\ \left(1 - r^5\right)\left(r^5 - 32\right) &= 0 \end{aligned}$$ $$r^5=1\ \text{(reject since $r\ne1$)}\quad\text{or}\quad r^5=32$$ $$r=\sqrt[5]{32}=2\quad\text{(shown)}$$

(aiii)

$u_6 \leq 11$ and $u_7 > 11$ $$\begin{aligned} a(2)^5 \leq 11 \quad&\text{and}\quad a(2)^6 > 11\\ 32a \leq 11 \quad&\text{and}\quad 64a > 11\\ \frac{11}{64} < a &\leq \frac{11}{32}. \end{aligned}$$

(b)

$$\begin{aligned} \sum_{r=4}^{n}\frac{1}{(2r+3)(2r+5)} &= \sum_{\substack{r-2=4}}^{\substack{r-2=n}}\frac{1}{(2r-1)(2r+1)} &&\text{(replace $r$ with $r-2$)}\\ &= \sum_{r=6}^{n+2}\frac{1}{(2r-1)(2r+1)}\\ &= \sum_{r=1}^{n+2}\frac{1}{(2r-1)(2r+1)} - \sum_{r=1}^{5}\frac{1}{(2r-1)(2r+1)}\\ &= \left(\frac{1}{2} - \frac{1}{4n+10}\right) - \left(\frac{1}{2} - \frac{1}{22}\right)\\ &= \frac{1}{22} - \frac{1}{4n+10}. \end{aligned}$$