Sequences & series (arithmetic, geometric, sigma, recurrence) – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

The third term of an arithmetic series is 17 and the eighth term is 42.

(a) Find the first term and the common difference of the series. [2]

(b) Find the sum of the first 20 terms of the series. [2]

2CoreCalculator5 marks

The second term of a geometric series is 24 and the fifth term is $-3$.

(a) Find the common ratio of the series. [2]

(b) Find the first term of the series. [1]

(a) Forms $ar=24,\ ar^{4}=-3$ and divides to obtain $r^{3}=-\tfrac18$M1
$r=-\tfrac12$A1
(b) $a=-48$B1
(c) Uses $S_\infty=\dfrac{a}{1-r}$ with their $a$ and $|r|<1$M1
$-32$A1
Answer: (a) $-\tfrac12$ (b) $-48$ (c) $-32$
Examiner tip: With a negative ratio, $1-r=1-(-\tfrac12)=\tfrac32$ – sign slips here are the usual lost A mark.

3StandardCalculator5 marks

Find the value of

(a) $\displaystyle\sum_{r=1}^{40}(3r-2)$ [2]

(b) $\displaystyle\sum_{r=5}^{12}2^{r}$ [3]

(a) Recognises an arithmetic series with $a=1,\ d=3$ (or $l=118$) and $n=40$ and uses a correct sum formula, e.g. $\tfrac{40}{2}(1+118)$M1
$2380$A1
(b) Identifies a geometric series with first term $2^{5}=32$ and $r=2$ (or writes $\sum_{1}^{12}-\sum_{1}^{4}$)M1
Uses $S_n=\dfrac{a(r^{n}-1)}{r-1}$ with $n=8$ (or $2^{13}-2-(2^{5}-2)$)M1
$8160$A1
Answer: (a) $2380$ (b) $8160$
Examiner tip: Count the terms from $r=5$ to $r=12$ carefully: there are 8, not 7.

4StandardCalculator6 marks

A sequence is defined by $$u_{1}=2,\qquad u_{n+1}=ku_{n}+3,\quad n\geqslant1$$ where $k$ is a constant.

(a) Find, in terms of $k$, an expression for $u_{3}$. [2]

Given that $u_{3}=5$,

(b) find the possible values of $k$. [2]

For the positive value of $k$ found in (b), the sequence converges to a limit $L$.

(a) Uses the recurrence twice: $u_{2}=2k+3$, $u_{3}=k(2k+3)+3$M1
$u_{3}=2k^{2}+3k+3$A1
(b) Sets $2k^{2}+3k+3=5$ and solves the 3TQ $2k^{2}+3k-2=0$M1
$k=\tfrac12$ or $k=-2$A1
(c) Sets $L=\tfrac12L+3$M1
$L=6$A1
Answer: (a) $u_3=2k^{2}+3k+3$ (b) $k=\tfrac12$ or $k=-2$ (c) $L=6$
Examiner tip: For a limit, replace both $u_{n+1}$ and $u_{n}$ by the same $L$.

5HardCalculator7 marks

A company's annual profit is modelled as increasing by 4% each year. In Year 1 its profit is £250 000.

(a) Show that the model predicts a profit of £270 400 in Year 3. [1]

(b) Find the first year in which the model predicts the annual profit will exceed £400 000. [3]

(c) Find the total profit predicted by the model for Years 1 to 15 inclusive, giving your answer to the nearest £1000. [2]

(d) Give a reason why the model may not be suitable for predicting the profit in Year 40. [1]

(a) $250\,000\times1.04^{2}=270\,400$ *B1
(b) Sets $250\,000\times1.04^{n-1}>400\,000$ (allow $=$; allow $n$ for $n-1$ for this mark). Alt (trial): finds profit for Year 12 or 13 (awrt £385 000 / £400 000) M1M1
Uses logs correctly: $n-1>\dfrac{\ln1.6}{\ln1.04}\ (=11.98\ldots)$. Alt (trial): finds both Year 12 ≈ £384 864 (< 400 000) and Year 13 ≈ £400 258 (> 400 000) M1M1
Year 13 (cao)A1
(c) Uses $S_{15}=\dfrac{250\,000(1.04^{15}-1)}{1.04-1}$M1
£5 006 000A1
(d) Any sensible reason, e.g. profit growth is unlikely to stay at a constant 4% for 40 years / economic conditions changeB1
Answer: (a) $250\,000\times1.04^2=270\,400$ (b) Year 13 (c) £5 006 000 (d) e.g. growth rate unlikely to stay constant
Examiner tip: The $n$th term uses $1.04^{n-1}$: getting $n-1>11.98$ means $n-1=12$, so Year 13 – not Year 12.

6StretchCalculator7 marks

The first three terms of a geometric series are $$(k+4),\quad k,\quad (2k-15)$$ where $k$ is a constant.

(a) Show that $k^{2}-7k-60=0$. [2]

Given that the series is convergent,

(b) find the value of $k$, justifying your answer. [3]

(a) Uses a common ratio property, e.g. $\dfrac{k}{k+4}=\dfrac{2k-15}{k}$M1
$k^{2}=(k+4)(2k-15)=2k^{2}-7k-60\Rightarrow k^{2}-7k-60=0$ csoA1*
(b) Solves the quadratic: $(k-12)(k+5)=0$, $k=12$ or $k=-5$M1
Finds $r$ for at least one value: $k=12\Rightarrow r=\tfrac34$; $k=-5\Rightarrow r=5$M1
$k=12$ since $|r|=\tfrac34<1$ (and $k=-5$ rejected as $|r|>1$)A1
(c) $S_\infty=\dfrac{16}{1-\frac34}$ with first term $k+4=16$M1
$64$A1
Answer: (b) $k=12$ (c) $64$
Examiner tip: 'Justify' means you must give $r$ for both values and say which satisfies $|r|<1$.

7StandardCalculator4 marks

A geometric series has first term $a$ and common ratio $r$, where $r\neq1$.

Prove that the sum of the first $n$ terms of the series is $$S_n=\frac{a(1-r^{n})}{1-r}$$ [4]

Writes $S_n=a+ar+ar^{2}+\dots+ar^{n-1}$ (at least first two and last term)B1
Multiplies by $r$: $rS_n=ar+ar^{2}+\dots+ar^{n-1}+ar^{n}$M1
Subtracts to obtain $S_n-rS_n=a-ar^{n}$ (allow $rS_n-S_n=ar^{n}-a$); dependent on the previous M1dM1
Factorises $S_n(1-r)=a(1-r^{n})$ and divides by $(1-r)$ to reach the given result with a concluding statement, cso (no need to restate $r\neq1$, which is given; if $rS_n-S_n$ used, must show $\frac{a(r^n-1)}{r-1}=\frac{a(1-r^n)}{1-r}$)A1*
Answer: Proof as in mark scheme
Examiner tip: The last term is $ar^{n-1}$, not $ar^{n}$ – a wrong final term breaks the subtraction step.

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