Arithmetic and geometric sequences and series – 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.

3StretchCalculator7 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 $|r|<1$.

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4StandardCalculator4 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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