2019 H2 Maths Paper 1, Question 8

Let $S_{64}$ be the sum of the first $64$ terms of the arithmetic series with first term $a$ and common difference $2a$.

The $k$th term of the geometric series with first term $a$ and common ratio $2$ is $ar^{k - 1} = a(2)^{k - 1}$. Using the given equation,

Since $a \neq 0$, we divide both sides by $a$:

Since $4096 = 2^{12}$,

Comparing indices,

Since the sum of the first four terms is $0$ and $f \neq 0$,

Grouping in pairs,

Since $r \in \mathbb{R}$, $1 + r^{2} > 0$ for every $r$, so $1 + r^{2} \neq 0$. Hence

Note: $f$ itself is not constrained further by this equation — any nonzero real value of $f$ gives a valid series once $r = - 1$.

$\therefore$ $r = - 1$, and $f$ may be any nonzero real number.

With $r = - 1$, the sum of the first $n$ terms is

If $n$ is even, $( - 1)^{n} = 1$, so $S_{n} = 0$. If $n$ is odd, $( - 1)^{n} = - 1$, so

$\therefore$ $S_{n} = 0$ if $n$ is even, and $S_{n} = f$ if $n$ is odd.

Let the arithmetic series have first term $a < 0$ and common difference $d$. The sum of the first four terms is $14$:

The product of the first four terms is $0$:

Since $a \neq 0$ (it is negative), exactly one of $a + d$, $a + 2d$, $a + 3d$ must be $0$. We test each case against $\left( \text{*} \right)$ and the requirement $a < 0$.

Case $a + d = 0$, i.e. $d = - a$. Substituting into $\left( \text{*} \right)$:

which satisfies $a < 0$. Then $d = - a = 7$. Checking: the four terms are $- 7,\ 0,\ 7,\ 14$, with product $0$. ✓ (valid)

Case $a + 2d = 0$, i.e. $d = - \frac{a}{2}$. Substituting into $\left( \text{*} \right)$:

which contradicts $a < 0$. (rejected)

Case $a + 3d = 0$, i.e. $d = - \frac{a}{3}$. Substituting into $\left( \text{*} \right)$:

which again contradicts $a < 0$. (rejected)

Only the first case is valid, so $a = - 7$ and $d = 7$. The $11$th term is