Question
An arithmetic sequence has first term 60 and common difference -2.5.
(a) Given that the k th term of the sequence is zero, find the value of k . [2]
Let Sn denote the sum of the first n terms of the sequence.
(b) Find the maximum value of Sn . [3]
▶️Answer/Explanation
Ans:
(a) Since $u_1=60$ and $d=-2.5$, $u_k=62.5-2.5k$. Solving $62.5-2.5k=0$, we have $k=25$. (b) Since $S_n = \frac{n}{2}\left[2\left(60\right)-2.5\left(n-1\right)\right]$, from the graphing calculator, we have
| $n$ | $S_n$ |
|---|---|
| $23$ | $747.5$ |
| $24$ | $750$ |
| $25$ | $750$ |
| $26$ | $747.5$ |
Thus, the maximum value of $S_n$ is $750$, which occurs at $n=24$ and $n=25$.
Question
The sum of the first 16 terms of an arithmetic sequence is 212 and the fifth term is 8.
▶️Answer/Explanation
Markscheme
\({S_n} = \frac{n}{2}[2a + (n – 1)d]\)
\(212 = \frac{{16}}{2}(2a + 15d)\,\,\,\,\,( = 16a + 120d)\) A1
\({n^{th}}{\text{ term is }}a + (n – 1)d\)
\(8 = a + 4d\) A1
solving simultaneously: (M1)
\(d = 1.5,{\text{ }}a = 2\) A1
[4 marks]
\(\frac{n}{2}[4 + 1.5(n – 1)] > 600\) (M1)
\( \Rightarrow 3{n^2} + 5n – 2400 > 0\) (A1)
\( \Rightarrow n > 27.4…,{\text{ }}(n < – 29.1…)\)
Note: Do not penalize improper use of inequalities.
\( \Rightarrow n = 28\) A1
[3 marks]
Examiners report
This proved to be a good start to the paper for most candidates. The vast majority made a meaningful attempt at this question with many gaining the correct answers. Candidates who lost marks usually did so because of mistakes in the working. In part (b) the most efficient way of gaining the answer was to use the calculator once the initial inequality was set up. A small number of candidates spent valuable time unnecessarily manipulating the algebra before moving to the calculator.
This proved to be a good start to the paper for most candidates. The vast majority made a meaningful attempt at this question with many gaining the correct answers. Candidates who lost marks usually did so because of mistakes in the working. In part (b) the most efficient way of gaining the answer was to use the calculator once the initial inequality was set up. A small number of candidates spent valuable time unnecessarily manipulating the algebra before moving to the calculator.
Question
Each time a ball bounces, it reaches 95 % of the height reached on the previous bounce.
Initially, it is dropped from a height of 4 metres.
a.What height does the ball reach after its fourth bounce?[2]
b.How many times does the ball bounce before it no longer reaches a height of 1 metre?[3]
c.What is the total distance travelled by the ball?[3]
▶️Answer/Explanation
Markscheme
height \( = 4 \times {0.95^4}\) (A1)
= 3.26 (metres) A1
[2 marks]
\(4 \times {0.95^n} < 1\) (M1)
\({0.95^n} < 0.25\)
\( \Rightarrow n > \frac{{\ln 0.25}}{{\ln 0.95}}\) (A1)
\( \Rightarrow n > 27.0\)
Note: Do not penalize improper use of inequalities.
\( \Rightarrow n = 28\) A1
Note: If candidates have used n – 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c).
[3 marks]
METHOD 1
recognition of geometric series with sum to infinity, first term of \(4 \times 0.95\) and common ratio 0.95 M1
recognition of the need to double this series and to add 4 M1
total distance travelled is \(2\left( {\frac{{4 \times 0.95}}{{1 – 0.95}}} \right) + 4 = 156{\text{ (metres)}}\) A1
[3 marks]
Note: If candidates have used n – 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c).
METHOD 2
recognition of a geometric series with sum to infinity, first term of 4 and common ratio 0.95 M1
recognition of the need to double this series and to subtract 4 M1
total distance travelled is \({\text{2}}\left( {\frac{4}{{1 – 0.95}}} \right) – 4 = 156{\text{ (metres)}}\) A1
[3 marks]
Question
Find the sum of all the multiples of 3 between 100 and 500.
▶️Answer/Explanation
Markscheme
METHOD 1
102 + 105 + … + 498 (M1)
so number of terms = 133 (A1)
EITHER
\( = \frac{{133}}{2}(2 \times 102 + 132 \times 3)\) (M1)
= 39900 A1
OR
\( = (102 + 498) \times \frac{{133}}{2}\) (M1)
= 39900 A1
OR
\(\sum\limits_{n = 34}^{166} {3n} \) (M1)
= 39900 A1
METHOD 2
\(500 \div 3 = 166.666…{\text{ and }}100 \div 3 = 33.333…\)
\(102 + 105 + … + 498 = \sum\limits_{n = 1}^{166} {3n} – \sum\limits_{n = 1}^{33} {3n} \) (M1)
\(\sum\limits_{n = 1}^{166} {3n} = 41583\) (A1)
\(\sum\limits_{n = 1}^{33} {3n} = 1683\) (A1)
the sum is 39900 A1
[4 marks]
Examiners report
Most candidates got full marks in this question. Some mistakes were detected when trying to find the number of terms of the arithmetic sequence, namely the use of the incorrect value n = 132 ; a few interpreted the question as the sum of multiples between the 100th and 500th terms. Occasional application of geometric series was attempted.
Question
A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence.
The angle of the largest sector is twice the angle of the smallest sector.
Find the size of the angle of the smallest sector.
▶️Answer/Explanation
Ans
METHOD 1
If the areas are in arithmetic sequence, then so are the angles.
\(Rightarrow S_n=\frac {n}{2}(a+1)\Rightarrow \frac {12}{2}(\theta +2\theta )=18\theta \)\
\(\Rightarrow 18\theta = 2\pi\)
\(\theta =\frac {\pi}{9}\) (accept 20o)
METHOD 2
\(a_12=2a_1\)
\(\frac {12}{2}(a_1+2a_1)=\pi r^2\)
\(3a_1=\frac {\pi r^2}{6}\)
\(\frac {3}{2}r^2\theta =\frac {\pi r^2}{6}\)
\(\theta =\frac {2\pi}{18}=\frac {\pi}{9}\)
METHOD 3
Let smallest angle = a, common difference = d
a + 11d = 2a
a = 11d
\(S_n=\frac {12}{2}(2a+11d)=2\pi\)
\(6(2a+a)=2\pi \)
\(18a=2\pi\)
\(a=\frac {\pi}{9}\) (accept 20o)
Question
A metal rod 1 metre long is cut into 10 pieces, the lengths of which form a geometric sequence. The length of the longest piece is 8 times the length of the shortest piece. Find, to the nearest millimetre, the length of the shortest piece.
▶️Answer/Explanation
Markscheme
the pieces have lengths \(a,{\text{ }}ar,{\text{ …, }}a{r^9}\) (M1)
\(8a = a{r^9}{\text{ }}({\text{or }}8 = {r^9})\) A1
\(r = \sqrt[9]{8} = 1.259922…\) A1
\(a\frac{{{r^{10}} – 1}}{{r – 1}} = 1\,\,\,\,\,\left( {{\text{or }}a\frac{{{r^{10}} – 1}}{{r – 1}} = 1000} \right)\) M1
\(a = \frac{{r – 1}}{{{r^{10}} – 1}} = 0.0286…\,\,\,\,\,\left( {{\text{or }}a = \frac{{r – 1}}{{{r^{10}} – 1}} = 28.6…} \right)\) (A1)
a = 29 mm (accept 0.029 m or any correct answer regardless the units) A1
[6 marks]
Question
A geometric sequence has a first term of 2 and a common ratio of 1.05. Find the value of the smallest term which is greater than 500.
▶️Answer/Explanation
Markscheme
\(2 \times {1.05^{n – 1}} > 500\) M1
\(n – 1 > \frac{{\log {\text{ }}250}}{{\log {\text{ }}1.05}}\) M1
\(n – 1 > 113.1675…\) A1
\(n = 115\) (A1)
\({u_{115}} = 521\) A1 N5
Note: Accept graphical solution with appropriate sketch.
[5 marks]
Examiners report
Many candidates misread the question and stopped at showing that the required term was the \({115^{{\text{th}}}}\).
Question
In the arithmetic series with \({n^{{\text{th}}}}\) term \({u_n}\) , it is given that \({u_4} = 7\) and \({u_9} = 22\) .
Find the minimum value of n so that \({u_1} + {u_2} + {u_3} + … + {u_n} > 10\,000\) .
▶️Answer/Explanation
Markscheme
\({u_4} = {u_1} + 3d = 7\) , \({u_9} = {u_1} + 8d = 22\) A1A1
Note: 5d = 15 gains both above marks
\({u_1} = – 2\) , \(d = 3\) A1
\({S_n} = \frac{n}{2}\left( { – 4 + (n – 1)3} \right) > 10\,000\) M1
\(n = 83\) A1
[5 marks]
Examiners report
This question was well answered by most candidates. A few did not realise that the answer had to be an integer.