2024 H2 Maths Paper 1 Q5 — A Geometric Series with a Negative Ratio

The series, livechoosing the ratio

Terms taken,

k

The terms

u_{r}

The head

S_{k}

The tail

T_{k}

Part (a)

3 marks

A geometric series has first term $a$ and common ratio $r$ , where $r\lt0$ . Its sum to infinity is 18 and its third term is $\frac{8}{3}$ . Show that $27r^{3}-27r^{2}+4=0$ .

Approach

Two facts, two unknowns. Write each as an equation, then eliminate $a$ — the question wants an equation in $r$ alone, so $a$ is the one that has to go.

The sum to infinity of a geometric series (valid since $\left|r\right|\lt1$ , which the existence of $S_{\infty}$ presupposes):

\frac{a}{1-r}=18\ \Longrightarrow\ a=18(1-r)\qquad\cdots\ (1)

The third term is $ar^{2}$ :

ar^{2}=\frac{8}{3}\qquad\cdots\ (2)

Substituting (1) into (2):

18(1-r)r^{2}=\frac{8}{3}

Multiplying both sides by 3 and expanding:

\begin{aligned}54(1-r)r^{2}&=8\\54r^{2}-54r^{3}&=8\\-54r^{3}+54r^{2}-8&=0\end{aligned}

Multiplying by $-1$ and dividing by 2:

27r^{3}-27r^{2}+4=0\quad\textbf{(shown)}

Where the marks go

The third term is $ar^{2}$ , not $ar^{3}$ — the $n$ th term is $ar^{n-1}$ , and off-by-one here poisons everything downstream. In a show question the target form is also part of the answer: finish by dividing by 2 so the constant reads 4, rather than leaving $54r^{3}-54r^{2}+8=0$ and hoping.

Part (b)

1 mark

Find the values of $r$ and $a$ .

Approach

One mark for a cubic, three roots and two conditions to apply — which tells you the factorisation is meant to be quick. Try the simple negative fractions first: the condition $r\lt0$ says where to look.

Testing $r=-\frac{1}{3}$ :

27\left(-\tfrac{1}{27}\right)-27\left(\tfrac{1}{9}\right)+4=-1-3+4=0

so $(3r+1)$ is a factor. Dividing:

27r^{3}-27r^{2}+4=(3r+1)\left(9r^{2}-12r+4\right)=(3r+1)(3r-2)^{2}

The roots are therefore $r=-\frac{1}{3}$ and $r=\frac{2}{3}$ (repeated). Since the question states $r\lt0$ :

r=-\frac{1}{3}

and from (1),

a=18\left(1-\left(-\tfrac{1}{3}\right)\right)=18\left(\tfrac{4}{3}\right)=24

To check your answer:

Third term $=ar^{2}=24\left(\frac{1}{9}\right)=\frac{8}{3}$ ✓, and $S_{\infty}=\frac{24}{1-\left(-\frac{1}{3}\right)}=\frac{24}{4/3}=18$ ✓. Both given facts recovered.

$\therefore\ r=-\dfrac{1}{3}$ and $a=24$ .

Note — the repeated root is not a coincidence

$(3r-2)^{2}$ means the cubic touches the $r$ -axis at $r=\frac{2}{3}$ rather than crossing it. Both $r=-\frac{1}{3}$ and $r=\frac{2}{3}$ give a genuine convergent series with $S_{\infty}=18$ and third term $\frac{8}{3}$ — the condition $r\lt0$ is not tidying up an impossible case, it is choosing between two real ones. Without it the question would have no single answer.

Part (c)

3 marks

The $n$ th term of the series is $u_{n}$ . Given that $\displaystyle 80\sum_{r=k+1}^{\infty}u_{r}=\sum_{r=1}^{k}u_{r}$ , find the value of $k$ .

Approach

The right-hand sum is the first $k$ terms — the head. The left-hand sum is everything after them — the tail. The two together make the whole infinite series, so the tail is simply $S_{\infty}-S_{k}$ and never needs to be summed from scratch.

Write $S_{k}=\displaystyle\sum_{r=1}^{k}u_{r}=\frac{a\left(1-r^{k}\right)}{1-r}$ and $T_{k}=\displaystyle\sum_{r=k+1}^{\infty}u_{r}$ . Since $S_{\infty}=S_{k}+T_{k}$ ,

T_{k}=S_{\infty}-S_{k}=\frac{a}{1-r}-\frac{a\left(1-r^{k}\right)}{1-r}=\frac{ar^{k}}{1-r}

The given condition $80\,T_{k}=S_{k}$ becomes

80\left(\frac{ar^{k}}{1-r}\right)=\frac{a\left(1-r^{k}\right)}{1-r}

Both sides carry the factor $\frac{a}{1-r}$ , which is $18\neq0$ , so it divides out:

\begin{aligned}80r^{k}&=1-r^{k}\\81r^{k}&=1\\r^{k}&=\frac{1}{81}\end{aligned}

With $r=-\frac{1}{3}$ :

\left(-\tfrac{1}{3}\right)^{k}=\frac{1}{81}

The right-hand side is positive, so $k$ must be even; and then $\left(\frac{1}{3}\right)^{k}=\frac{1}{81}=\left(\frac{1}{3}\right)^{4}$ , giving $k=4$ , which is indeed even.

To check your answer:

$S_{4}=\frac{24\left(1-\frac{1}{81}\right)}{4/3}=\frac{160}{9}$ , so $T_{4}=18-\frac{160}{9}=\frac{2}{9}$ , and $80\times\frac{2}{9}=\frac{160}{9}=S_{4}$ ✓. Slide $k$ in the model: the ratio head : tail passes 80 at $k=4$ and nowhere else.

$\therefore\ k=4$ .

Where the marks go

Summing the tail directly — first term $ar^{k}$ , ratio $r$ , so $\frac{ar^{k}}{1-r}$ — is perfectly valid and lands in the same place, but the first term of the tail is $u_{k+1}=ar^{k}$ , not $ar^{k+1}$ , and that is where it goes wrong. The other loss is finishing at $\left(\frac{1}{3}\right)^{k}=\frac{1}{81}$ without saying why the negative sign is allowed to disappear: $k$ even is a step in the argument, not a detail.

What this question was really testing

Eliminating the right unknown. Two equations, two unknowns, but the target is an equation in $r$ — so $a$ is the one to substitute away, not the one to solve for.
Using the given condition to choose a root. $r\lt0$ is not decoration; the cubic has two valid ratios and the condition picks one.
Head plus tail equals the whole. $\sum_{k+1}^{\infty}=S_{\infty}-S_{k}$ is the single idea in part (c), and it turns a nasty-looking sum into one line.
Parity of an exponent. When a negative ratio is raised to an unknown power and the result is positive, the exponent must be even. Saying so is a mark.

A series that overshoots, then corrects, then overshoots less

Part (a)

Part (b)

Part (c)

What this question was really testing

This is how we mark, and how we teach.