Arithmetic and geometric progressions: SAJC 2024 Promo Q5

(a)

For the sum to infinity to exist, $$\begin{aligned} |2\sin\theta| &< 1 \quad\text{(*)}\\ -1 < 2\sin\theta &< 1\\ -\frac{1}{2} < \sin\theta &< \frac{1}{2}\\ -\frac{\pi}{6} < \theta &< \frac{\pi}{6}. \quad\text{(1)} \end{aligned}$$ For the sum to infinity to be greater than $2$, $$\begin{aligned} \frac{1}{1 - 2\sin\theta} &> 2. \quad\text{(\#)} \end{aligned}$$ $$\begin{aligned} 0<1-2\sin\theta&<\frac12\qquad\text{since }|2\sin\theta|<1,\\ \sin\theta>\frac14&\Rightarrow\theta>0.253. \quad\text{(2)} \end{aligned}$$ Since $-\dfrac{\pi}{6}<\theta<\dfrac{\pi}{6}$, therefore $\sin^{-1}(\frac14)<\theta<\frac\pi6$ (final answer).

Alternatively, students can graph $y=\dfrac{1}{1-2\sin\theta}$.

From the graph, $\dfrac{1}{1-2\sin\theta}>2$. $$\Rightarrow \sin^{-1}(\frac14)<\theta<\frac\pi6\quad\text{(final answer)}.$$

(b)

Let $a$ denote the first term and $r$ be the common ratio of the geometric progression respectively.

Likewise, let $b$ and $d$ denote the first term and common difference of the arithmetic progression. Then $$\begin{aligned} ar^4 &= b + 6d \quad\text{(1)}\\ ar^8 &= b + 24d \quad\text{(2)}\\ ar^{10} &= b + 49d. \quad\text{(3)} \end{aligned}$$ Taking $(2) - (1)$ and $(3) - (2)$: $$\begin{aligned} ar^8 - ar^4 &= 18d \quad\text{(4)}\\ ar^{10} - ar^8 &= 25d. \quad\text{(5)} \end{aligned}$$ Dividing $(5)$ by $(4)$: $$\begin{aligned} \frac{ar^8(r^2 - 1)}{ar^4(r^4 - 1)} &= \frac{25}{18}\\ \frac{r^4(r^2 - 1)}{(r^2 + 1)(r^2 - 1)} &= \frac{25}{18}\\ \frac{r^4}{r^2 + 1} &= \frac{25}{18}\\ 18r^4 &= 25r^2 + 25\\ 18r^4 - 25r^2 - 25 &= 0. \quad\text{((shown))} \end{aligned}$$ Solving for $r^2$: $$\begin{aligned} r^2 &= \frac{25 \pm \sqrt{(-25)^2 - 4(18)(-25)}}{2(18)} = \frac{25 \pm \sqrt{2425}}{36}. \end{aligned}$$

Taking the positive value (since $r^2 > 0$), $$\begin{aligned} r &= \pm\sqrt{\frac{25 + \sqrt{2425}}{36}} = 1.436 \text{ or } -1.436. \end{aligned}$$ For both values, $|r| = 1.436 > 1$, so the geometric progression is not convergent.

Alternatively, students can use GC to solve $18r^4-25r^2-25=0$.

The 4 roots are $1.436$, $-1.436$, $0.821\mathrm i$, $-0.821\mathrm i$.

However we only need to consider real roots.

Hence $r=1.436$ or $-1.436$.

The solution’s upper endpoint has been corrected to π/6 (approximately 0.524 radians); the printed value 0.526 was inaccurate.