Arithmetic and geometric progressions: NYJC 2024 Promo Q6

(a)

NEW $$\begin{aligned} S_n&=n^2+2n\\S_n&=n^2+2n\\u_n&=S_n-S_{n-1}\\ &=\left(n^2+2n\right)-\left((n-1)^2+2(n-1)\right)\\ &=n^2+2n-n^2+2n-1-2n+2\\&=2n+1 \end{aligned}$$ $$\begin{aligned} u_{n+1}-u_n&=(2(n+1)+1)-(2n+1)\\&=2n+3-2n-1\\&=2 \end{aligned}$$ Since $u_{n+1}-u_n=2$ is a constant, it is an arithmetic series. $$\therefore d=2$$ $$\begin{aligned} u_n&=2n+1\\a+(n-1)2&=2n+1\\a+2n-2&=2n+1\\a&=3 \end{aligned}$$

(b)

$$\begin{aligned} a&=12\\u_6&=-\frac38\\\Rightarrow ar^5&=-\frac38\\ \Rightarrow r^5&=-\frac1{32}\\\Rightarrow r&=-\frac12 \end{aligned}$$ $$\begin{aligned} S_n&=\frac{a(1-r^n)}{1-r}\\ &=\frac{12\left(1-\left(-\frac12\right)^n\right)}{1-\left(-\frac12\right)}\\ &=8\left(1-\left(-\frac12\right)^n\right) \end{aligned}$$ $$\begin{aligned} S&=\frac{a}{1-r}\\&=\frac{12}{1-\left(-\frac12\right)}\\&=8 \end{aligned}$$ $$\begin{aligned} \lvert S_n-S\rvert&<0.001\\ \left\lvert8\left(1-\left(-\frac12\right)^n\right)-8\right\rvert&<0.001\\ \left\lvert8\left(1-\left(-\frac12\right)^n\right)-8\right\rvert-0.001&<0 \end{aligned}$$ From GC,

Therefore, the least value of $n$ is 13.