Question 3. [Maximum mark: 5]
An arithmetic sequence has first term 60 and common difference -2.5.
(a) Given that the kth term of the sequence is zero, find the value of k . [2]
Let \(S_{n}\) denote the sum of the first n terms of the sequence.
(b) Find the maximum value of \(S_{n}\)
Answer/Explanation
(a) attempt to use \(u_{1}+\left ( n-1 \right )d=0\)
\(60-2.5\left ( k-1 \right )=0\)
k=25
(b) METHOD 1
attempting to express \(S_{n}\) in terms of n use of graph or a table to attempt to find the maximum sum =750
EITHER
recogni9zing maximum occurs at n=25
\(S_{25}=\frac{25}{2}\left ( 60+0 \right ),S_{25}=\frac{25}{2}\left (2\times 60+24\times -2.5 \right )\)
THEN
=750
Question
A city is concerned about pollution, and decides to look at the number of people using taxis. At the end of the year 2000, there were 280 taxis in the city. After n years the number of taxis, T, in the city is given by\[T = 280 \times {1.12^n} .\]
(i) Find the number of taxis in the city at the end of 2005.
(ii) Find the year in which the number of taxis is double the number of taxis there were at the end of 2000.
At the end of 2000 there were \(25600\) people in the city who used taxis.
After n years the number of people, P, in the city who used taxis is given by\[P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ – 0.1n}}}} .\](i) Find the value of P at the end of 2005, giving your answer to the nearest whole number.
(ii) After seven complete years, will the value of P be double its value at the end of 2000? Justify your answer.
Let R be the ratio of the number of people using taxis in the city to the number of taxis. The city will reduce the number of taxis if \(R < 70\) .
(i) Find the value of R at the end of 2000.
(ii) After how many complete years will the city first reduce the number of taxis?
Answer/Explanation
Markscheme
(i) \(n = 5\) (A1)
\(T = 280 \times {1.12^5}\)
\(T = 493\) A1 N2
(ii) evidence of doubling (A1)
e.g. 560
setting up equation A1
e.g. \(280 \times {1.12^n} = 560\), \({1.12^n} = 2\)
\(n = 6.116 \ldots \) (A1)
in the year 2007 A1 N3
[6 marks]
(i) \(P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ – 0.1(5)}}}}\) (A1)
\(P = 39635.993 \ldots \) (A1)
\(P = 39636\) A1 N3
(ii) \(P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ – 0.1(7)}}}}\)
\(P = 46806.997 \ldots \) A1
not doubled A1 N0
valid reason for their answer R1
e.g. \(P < 51200\)
[6 marks]
(i) correct value A2 N2
e.g. \(\frac{{25600}}{{280}}\) , 91.4, \(640:7\)
(ii) setting up an inequality (accept an equation, or reversed inequality) M1
e.g. \(\frac{P}{T} < 70\) , \(\frac{{2560000}}{{(10 + 90{{\rm{e}}^{ – 0.1n}})280 \times {{1.12}^n}}} < 70\)
finding the value \(9.31 \ldots \) (A1)
after 10 years A1 N2
[5 marks]
Question
In an arithmetic series, the first term is –7 and the sum of the first 20 terms is 620.
Find the common difference.
Find the value of the 78th term.
Answer/Explanation
Markscheme
attempt to substitute into sum formula for AP (accept term formula) (M1)
e.g. \({S_{20}} = \frac{{20}}{2}\left\{ {2( – 7) + 19\left. d \right\}} \right.\) , \(\left( {{\rm{or}}\frac{{20}}{2}\left( { – 7 + {u_{20}}} \right)} \right)\)
setting up correct equation using sum formula A1
e.g. \(\frac{{20}}{2}\left\{ {2( – 7) + \left. {19d} \right\}} \right. = 620\)
\(d = 4\) A1 N2
[3 marks]
correct substitution \({u_{78}} = – 7 + 77(4)\) (A1)
= 301 A1 N2
[2 marks]
Question
In an arithmetic sequence, \({S_{40}} = 1900\) and \({u_{40}} = 106\) . Find the value of \({u_1}\) and of d .
Answer/Explanation
Markscheme
METHOD 1
substituting into formula for \({S_{40}}\) (M1)
correct substitution A1
e.g. \(1900 = \frac{{40({u_1} + 106)}}{2}\)
\({u_1} = – 11\) A1 N2
substituting into formula for \({u_{40}}\) or \({S_{40}}\) (M1)
correct substitution A1
e.g. \(106 = – 11 + 39d\) , \(1900 = 20( – 22 + 39d)\)
\(d = 3\) A1 N2
METHOD 2
substituting into formula for \({S_{40}}\) (M1)
correct substitution A1
e.g. \(20(2{u_1} + 39d) = 1900\)
substituting into formula for \({u_{40}}\) (M1)
correct substitution A1
e.g. \(106 = {u_1} + 39d\)
\({u_1} = – 11\) , \(d = 3\) A1A1 N2N2
[6 marks]
Question
Consider the arithmetic sequence 3, 9, 15, \(\ldots \) , 1353 .
Write down the common difference.
Find the number of terms in the sequence.
Find the sum of the sequence.
Answer/Explanation
Markscheme
common difference is 6 A1 N1
[1 mark]
evidence of appropriate approach (M1)
e.g. \({u_n} = 1353\)
correct working A1
e.g. \(1353 = 3 + (n – 1)6\) , \(\frac{{1353 + 3}}{6}\)
\(n = 226\) A1 N2
[3 marks]
evidence of correct substitution A1
e.g. \({S_{226}} = \frac{{226(3 + 1353)}}{2}\) , \(\frac{{226}}{2}(2 \times 3 + 225 \times 6)\)
\({S_{226}} = 153228\) (accept 153000) A1 N1
[2 marks]
Question
The nth term of an arithmetic sequence is given by \({u_n} = 5 + 2n\) .
Write down the common difference.
(i) Given that the nth term of this sequence is 115, find the value of n .
(ii) For this value of n , find the sum of the sequence.
Answer/Explanation
Markscheme
\(d = 2\) A1 N1
[1 mark]
(i) \(5 + 2n = 115\) (A1)
\(n = 55\) A1 N2
(ii) \({u_1} = 7\) (may be seen in above) (A1)
correct substitution into formula for sum of arithmetic series (A1)
e.g. \({S_{55}} = \frac{{55}}{2}(7 + 115)\) , \({S_{55}} = \frac{{55}}{2}(2(7) + 54(2))\) , \(\sum\limits_{k = 1}^{55} {(5 + 2k)} \)
\({S_{55}} = 3355\) (accept \(3360\)) A1 N3
[5 marks]