Arithmetic and geometric progressions: CJC 2024 Promo Q8

(a)

$$\begin{aligned} S_{36}&\geq500\\ \frac{36}{2}\left[2a + (36 - 1)(0.5)\right] &\geq 500\\ \end{aligned}$$ Method 1⃝: Algebraic $$\begin{aligned} \frac{36}{2}\left[2a + (36 - 1)(0.5)\right] &\geq 500\\ 18(2a + 17.5) &\geq 500\\ 2a + 17.5 &\geq \frac{500}{18}\\ 2a &\geq \frac{500}{18} - 17.5\\ a &\geq 5.138888889. \end{aligned}$$ Hence, smallest value of $a$ is $6$.

Method 2⃝: G.C. Table

Hence, smallest value of $a$ is $6$.

(b)

(i) Method 1⃝: $$\begin{aligned} u_n &= 1.01u_{n-1} - x\\ &=1.01(1.01u_{n-2}-x)-x\\ &=1.01^2u_{n-2}-1.01x-x\\ &=1.01^2(1.01u_{n-2}-x)-1.01x-x\\ &=1.01^3u_{n-3}-1.01^2x-1.01x-x\\ &=\cdots\\ &= 1.01^n u_0 - 1.01^{n-1}x - \cdots - 1.01^2x - 1.01x - x\\ &= 1.01^n(400) - x\left(1 + 1.01 + 1.01^2 + \cdots + 1.01^{n-1}\right)\\ &= 1.01^n(400) - x\left[\frac{1(1.01^n - 1)}{1.01 - 1}\right]\\ &= 1.01^n(400) - 100x(1.01^n - 1). \quad\text{((shown))} \end{aligned}$$

Method 2⃝: $$\begin{aligned} u_1&=1.01u_0-x\\ u_2&=1.01u_1-x\\&=1.01(1.01u_0-x)-x\\&=1.01^2u_0-1.01x-x\\ u_3&=1.01u_2-x\\&=1.01(1.01^2u_0-1.01x-x)-x\\&=1.01^3u_0-1.01^2x-1.01x-x\\ u_n&=1.01^nu_0-1.01^{n-1}x-\cdots-1.01^2x-1.01x-x\\ &=1.01^n(400)-x(1+1.01+1.01^2+\cdots+1.01^{n-1})\\ &=1.01^n(400)-x\left[\frac{1(1.01^n-1)}{1.01-1}\right]\\ &=1.01^n(400)-100x(1.01^n-1)\quad\text{(shown)} \end{aligned}$$

(b)

(ii) $$\begin{aligned} y &= 1.01\left(1.01^k(400) - 100(16)(1.01^k - 1)\right)\\ &= 1.01^{k+1}(400) - 1616(1.01^k - 1). \end{aligned}$$ Since $0 < y < 16$, $$\begin{aligned} 0 < 1.01^{k+1}(400) - 1616(1.01^k - 1) < 16. \end{aligned}$$ Method 1⃝:

Using G.C.,

Hence, $k=28$.

Method 2⃝:

Using G.C.,

$$\begin{aligned} 27.91181&<k<28.91181. \end{aligned}$$ Hence, $k=28$.

Method 3⃝: $$\begin{aligned} 0&<1.01^{k+1}(400)-1616(1.01^k-1)<16\\ 0&<1.01^k(404)-1616(1.01^k)+1616<16\\ 0&<-1212(1.01^k)+1616<16\\ -1616&<-1212(1.01^k)<-1600\\ \frac{-1600}{-1212}&<1.01^k<\frac{-1616}{-1212}\\ \frac{400}{303}&<1.01^k<\frac43\\ \ln\left(\frac{400}{303}\right)&<k\ln1.01<\ln\left(\frac43\right)\\ 27.91180974&<k<28.91180974 \end{aligned}$$ Hence, $k=28$.

$$\begin{aligned} \therefore\ y &= 14.6\quad(\text{to 3 s.f.}) \end{aligned}$$