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Question
Consider the infinite geometric sequence \(3000{\text{, }}- 1800{\text{, }}1080{\text{, }} – 648, \ldots \) .
Find the common ratio.
Find the 10th term.
Find the exact sum of the infinite sequence.
Answer/Explanation
Markscheme
evidence of dividing two terms (M1)
e.g. \( – \frac{{1800}}{{3000}}\) , \( – \frac{{1800}}{{1080}}\)
\(r = – 0.6\) A1 N2
[2 marks]
evidence of substituting into the formula for the 10th term (M1)
e.g. \({u_{10}} = 3000{( – 0.6)^9}\)
\({u_{10}} = 30.2\) (accept the exact value \( – 30.233088\)) A1 N2
[2 marks]
evidence of substituting into the formula for the infinite sum (M1)
e.g. \(S = \frac{{3000}}{{1.6}}\)
\(S = 1875\) A1 N2
[2 marks]
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 a geometric series, \({u_1} = \frac{1}{{81}}\) and \({u_4} = \frac{1}{3}\) .
Find the value of \(r\) .
Find the smallest value of n for which \({S_n} > 40\) .
Answer/Explanation
Markscheme
evidence of substituting into formula for \(n\)th term of GP (M1)
e.g. \({u_4} = \frac{1}{{81}}{r^3}\)
setting up correct equation \(\frac{1}{{81}}{r^3} = \frac{1}{3}\) A1
\(r = 3\) A1 N2
[3 marks]
METHOD 1
setting up an inequality (accept an equation) M1
e.g. \(\frac{{\frac{1}{{81}}({3^n} – 1)}}{2} > 40\) , \(\frac{{\frac{1}{{81}}(1 – {3^n})}}{{ – 2}} > 40\) , \({3^n} > 6481\)
evidence of solving M1
e.g. graph, taking logs
\(n > 7.9888 \ldots \) (A1)
\(\therefore n = 8\) A1 N2
METHOD 2
if \(n = 7\) , sum \( = 13.49 \ldots \) ; if \(n = 8\) , sum \( = 40.49 \ldots \) A2
\(n = 8\) (is the smallest value) A2 N2
[4 marks]
Question
Expand \(\sum\limits_{r = 4}^7 {{2^r}} \) as the sum of four terms.
(i) Find the value of \(\sum\limits_{r = 4}^{30} {{2^r}} \) .
(ii) Explain why \(\sum\limits_{r = 4}^\infty {{2^r}} \) cannot be evaluated.
Answer/Explanation
Markscheme
\(\sum\limits_{r = 4}^7 {{2^r}} = {2^4} + {2^5} + {2^6} + {2^7}\) (accept \(16 + 32 + 64 + 128\) ) A1 N1
[1 mark]
(i) METHOD 1
recognizing a GP (M1)
\({u_1} = {2^4}\) , \(r = 2\) , \(n = 27\) (A1)
correct substitution into formula for sum (A1)
e.g. \({S_{27}} = \frac{{{2^4}({2^{27}} – 1)}}{{2 – 1}}\)
\({S_{27}} = 2147483632\) A1 N4
METHOD 2
recognizing \(\sum\limits_{r = 4}^{30} { = \sum\limits_{r = 1}^{30} { – \sum\limits_{r = 1}^3 {} } } \) (M1)
recognizing GP with \({u_1} = 2\) , \(r = 2\) , \(n = 30\) (A1)
correct substitution into formula for sum
\({S_{30}} = \frac{{2({2^{30}} – 1)}}{{2 – 1}}\) (A1)
\( = 214783646\)
\(\sum\limits_{r = 4}^{30} {{2^r}} = 2147483646 – (2 + 4 + 8)\)
\( = 2147483632\) A1 N4
(ii) valid reason (e.g. infinite GP, diverging series), and \(r \ge 1\) (accept \(r > 1\) ) R1R1 N2
[6 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
An arithmetic sequence, \({u_1}{\text{, }}{u_2}{\text{, }}{u_3} \ldots ,\) has \(d = 11\) and \({u_{27}} = 263\) .
Find \({u_1}\).
(i) Given that \({u_n} = 516\) , find the value of n .
(ii) For this value of n , find \({S_n}\) .
Answer/Explanation
Markscheme
evidence of equation for \({u_{27}}\) M1
e.g. \(263 = {u_1} + 26 \times 11\) , \({u_{27}} = {u_1} + (n – 1) \times 11\) , \(263 – (11 \times 26)\)
\({u_1} = – 23\) A1 N1
[2 marks]
(i) correct equation A1
e.g. \(516 = – 23 + (n – 1) \times 11\) , \(539 = (n – 1) \times 11\)
\(n = 50\) A1 N1
(ii) correct substitution into sum formula A1
e.g. \({S_{50}} = \frac{{50( – 23 + 516)}}{2}\) , \({S_{50}} = \frac{{50(2 \times ( – 23) + 49 \times 11)}}{2}\)
\({S_{50}} = 12325\) (accept 12300) A1 N1
[4 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]
Question
In an arithmetic sequence \({u_1} = 7\) , \({u_{20}} = 64\) and \({u_n} = 3709\) .
Find the value of the common difference.
Find the value of n .
Answer/Explanation
Markscheme
evidence of choosing the formula for 20th term (M1)
e.g. \({u_{20}} = {u_1} + 19d\)
correct equation A1
e.g. \(64 = 7 + 19d\) , \(d = \frac{{64 – 7}}{{19}}\)
\(d = 3\) A1 N2
[3 marks]
correct substitution into formula for \({u_n}\) A1
e.g. \(3709 = 7 + 3(n – 1)\) , \(3709 = 3n + 4\)
\(n = 1235\) A1 N1
[2 marks]
Question
Consider an infinite geometric sequence with \({u_1} = 40\) and \(r = \frac{1}{2}\) .
(i) Find \({u_4}\) .
(ii) Find the sum of the infinite sequence.
Consider an arithmetic sequence with n terms, with first term (\( – 36\)) and eighth term (\( – 8\)) .
(i) Find the common difference.
(ii) Show that \({S_n} = 2{n^2} – 38n\) .
The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n .
Answer/Explanation
Markscheme
(i) correct approach (A1)
e.g. \({u_4} = (40){\frac{1}{2}^{(4 – 1)}}\) , listing terms
\({u_4} = 5\) A1 N2
(ii) correct substitution into formula for infinite sum (A1)
e.g. \({S_\infty } = \frac{{40}}{{1 – 0.5}}\) , \({S_\infty } = \frac{{40}}{{0.5}}\)
\({S_\infty } = 80\) A1 N2
[4 marks]
(i) attempt to set up expression for \({u_8}\) (M1)
e.g. \( – 36 + (8 – 1)d\)
correct working A1
e.g. \( – 8 = – 36 + (8 – 1)d\) , \(\frac{{ – 8 – ( – 36)}}{7}\)
\(d = 4\) A1 N2
(ii) correct substitution into formula for sum (A1)
e.g. \({S_n} = \frac{n}{2}(2( – 36) + (n – 1)4)\)
correct working A1
e.g. \({S_n} = \frac{n}{2}(4n – 76)\) , \( – 36n + 2{n^2} – 2n\)
\({S_n} = 2{n^2} – 38n\) AG N0
[5 marks]
multiplying \({S_n}\) (AP) by 2 or dividing S (infinite GP) by 2 (M1)
e.g. \(2{S_n}\) , \(\frac{{{S_\infty }}}{2}\) , 40
evidence of substituting into \(2{S_n} = {S_\infty }\) A1
e.g. \(2{n^2} – 38n = 40\) , \(4{n^2} – 76n – 80\) (\( = 0\))
attempt to solve their quadratic (equation) (M1)
e.g. intersection of graphs, formula
\(n = 20\) A2 N3
[5 marks]
Question
The first three terms of an arithmetic sequence are 36, 40, 44,….
(i) Write down the value of d .
(ii) Find \({u_8}\) .
(i) Show that \({S_n} = 2{n^2} + 34n\) .
(ii) Hence, write down the value of \({S_{14}}\) .
Answer/Explanation
Markscheme
(i) \(d = 4\) A1 N1
(ii) evidence of valid approach (M1)
e.g. \({u_8} = 36 + 7(4)\) , repeated addition of d from 36
\({u_8} = 64\) A1 N2
[3 marks]
(i) correct substitution into sum formula A1
e.g. \({S_n} = \frac{n}{2}\left\{ {2\left( {36} \right) + (n – 1)(4)} \right\}\) , \(\frac{n}{2}\left\{ {72 + 4n – 4} \right\}\)
evidence of simplifying
e.g. \(\frac{n}{2}\left\{ {4n + 68} \right\}\) A1
\({S_n} = 2{n^2} + 34n\) AG N0
(ii) \(868\) A1 N1
[3 marks]
Question
The first term of a geometric sequence is 200 and the sum of the first four terms is 324.8.
Find the common ratio.
The first term of a geometric sequence is 200 and the sum of the first four terms is 324.8.
Find the tenth term.
Answer/Explanation
Markscheme
correct substitution into sum of a geometric sequence (A1)
e.g. \(200\left( {\frac{{1 – {r^4}}}{{1 – r}}} \right)\) , \(200 + 200r + 200{r^2} + 200{r^3}\)
attempt to set up an equation involving a sum and 324.8 M1
e.g. \(200\left( {\frac{{1 – {r^4}}}{{1 – r}}} \right) = 324.8\) , \(200 + 200r + 200{r^2} + 200{r^3} = 324.8\)
\(r = 0.4\) (exact) A2 N3
[4 marks]
correct substitution into formula A1
e.g. \({u_{10}} = 200 \times {0.4^9}\)
\({u_{10}} = 0.0524288\) (exact), \(0.0524\) A1 N1
[2 marks]
Question
The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .
Find the common difference.
The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .
Find the 28th term of the sequence.
The first three terms of an arithmetic sequence are \(5\) , \(6.7\) , \(8.4\) .
Find the sum of the first 28 terms.
Answer/Explanation
Markscheme
valid method (M1)
e.g. subtracting terms, using sequence formula
\(d = 1.7\) A1 N2
[2 marks]
correct substitution into term formula (A1)
e.g. \(5 + 27(1.7)\)
28th term is 50.9 (exact) A1 N2
[2 marks]
correct substitution into sum formula (A1)
e.g. \({S_{28}} = \frac{{28}}{2}(2(5) + 27(1.7))\) , \(\frac{{28}}{2}(5 + 50.9)\)
\({S_{28}} = 782.6\) (exact) [\(782\), \(783\)] A1 N2
[2 marks]
Question
An arithmetic sequence is given by \(5\), \(8\), \(11\), ….
(a) Write down the value of \(d\) .
(b) Find
(i) \({u_{100}}\) ;
(ii) \({S_{100}}\) .
(c) Given that \({u_n} = 1502\) , find the value of \(n\) .
Write down the value of \(d\) .
Find
(i) \({u_{100}}\) ;
(ii) \({S_{100}}\) .
Given that \({u_n} = 1502\) , find the value of \(n\) .
Answer/Explanation
Markscheme
(a) \(d = 3\) A1 N1
[1 mark]
(b) (i) correct substitution into term formula (A1)
e.g. \({u_{100}} = 5 + 3(99)\) , \(5 + 3(100 – 1)\)
\({u_{100}} = 302\) A1 N2
(ii) correct substitution into sum formula (A1)
eg \({S_{100}} = \frac{{100}}{2}(2(5) + 99(3))\) , \({S_{100}} = \frac{{100}}{2}(5 + 302)\)
\({S_{100}} = 15350\) A1 N2
[4 marks]
(c) correct substitution into term formula (A1)
eg \(1502 = 5 + 3(n – 1)\) , \(1502 = 3n + 2\)
\(n = 500\) A1 N2
[2 marks]
Total [7 marks].
\(d = 3\) A1 N1
[1 mark]
(i) correct substitution into term formula (A1)
e.g. \({u_{100}} = 5 + 3(99)\) , \(5 + 3(100 – 1)\)
\({u_{100}} = 302\) A1 N2
(ii) correct substitution into sum formula (A1)
eg \({S_{100}} = \frac{{100}}{2}(2(5) + 99(3))\) , \({S_{100}} = \frac{{100}}{2}(5 + 302)\)
\({S_{100}} = 15350\) A1 N2
[4 marks]
correct substitution into term formula (A1)
eg \(1502 = 5 + 3(n – 1)\) , \(1502 = 3n + 2\)
\(n = 500\) A1 N2
[2 marks]
Total [7 marks]
Question
The sum of the first three terms of a geometric sequence is \(62.755\), and the sum of the infinite sequence is \(440\). Find the common ratio.
Answer/Explanation
Markscheme
correct substitution into sum of a geometric sequence A1
eg \(62.755 = {u_1}\left( {\frac{{1 – {r^3}}}{{1 – r}}} \right)\) , \({u_1} + {u_1}r + {u_1}{r^2} = 62.755\)
correct substitution into sum to infinity A1
eg \(\frac{{{u_1}}}{{1 – r}} = 440\)
attempt to eliminate one variable (M1)
eg substituting \({u_1} = 440(1 – r)\)
correct equation in one variable (A1)
eg \(62.755 = 440(1 – r)\left( {\frac{{1 – {r^3}}}{{1 – r}}} \right)\) , \(440(1 – r)(1 + r + {r^2}) = 62.755\)
evidence of attempting to solve the equation in a single variable (M1)
eg sketch, setting equation equal to zero, \(62.755 = 440(1 – {r^3})\)
\(r =0.95 = \frac{{19}}{{20}}\) A1 N4
[6 marks]
Question
The first two terms of a geometric sequence \({u_n}\) are \({u_1} = 4\) and \({u_2} = 4.2\).
(i) Find the common ratio.
(ii) Hence or otherwise, find \({u_5}\).
Another sequence \({v_n}\) is defined by \({v_n} = a{n^k}\), where \(a,{\text{ }}k \in \mathbb{R}\), and \(n \in {\mathbb{Z}^ + }\), such that \({v_1} = 0.05\) and \({v_2} = 0.25\).
(i) Find the value of \(a\).
(ii) Find the value of \(k\).
Find the smallest value of \(n\) for which \({v_n} > {u_n}\).
Answer/Explanation
Markscheme
(i) valid approach (M1)
eg\(\;\;\;\)\(r = \frac{{{u_2}}}{{{u_1}}},{\text{ }}\frac{4}{{4.2}}\)
\(r = 1.05\;\;\;{\text{(exact)}}\) A1 N2
(ii) attempt to substitute into formula, with their \(r\) (M1)
eg\(\;\;\;\)\(4 \times {1.05^n},{\text{ }}4 \times 1.05 \times 1.05 \ldots \)
correct substitution (A1)
eg\(\;\;\;\)\(4 \times {1.05^4},{\text{ }}4 \times 1.05 \times 1.05 \times 1.05 \times 1.05\)
\({u_5} = 4.862025{\text{ (exact), }}4.86{\text{ }}[4.86,{\text{ }}4.87]{\text{ }}\) A1 N2
[5 marks]
(i) attempt to substitute \(n = 1\) (M1)
eg\(\;\;\;\)\(0.05 = a \times {1^k}\)
\(a = 0.05\) A1 N2
(ii) correct substitution of \(n = 2\) into \({v_2}\) A1
eg\(\;\;\;\)\(0.25 = a \times {2^k}\)
correct work (A1)
eg\(\;\;\;\)finding intersection point, \(k = {\log _2}\left( {\frac{{0.25}}{{0.05}}} \right),{\text{ }}\frac{{\log 5}}{{\log 2}}\)
\(2.32192\)
\(k = {\log _2}5\;\;\;{\text{(exact), }}2.32{\text{ }}[2.32,{\text{ }}2.33]\) A1 N2
[5 marks]
correct expression for \({u_n}\) (A1)
eg\(\;\;\;\)\(4 \times {1.05^{n – 1}}\)
EITHER
correct substitution into inequality (accept equation) (A1)
eg\(\;\;\;\)\(0.05 \times {n^k} > 4 \times {1.05^{n – 1}}\)
valid approach to solve inequality (accept equation) (M1)
eg\(\;\;\;\)finding point of intersection, \(n = 7.57994{\text{ }}(7.59508{\text{ from 2.32)}}\)
\(n = 8\;\;\;\)(must be an integer) A1 N2
OR
table of values
when \(n = 7,{\text{ }}{u_7} = 5.3604,{\text{ }}{v_7} = 4.5836\) A1
when \(n = 8,{\text{ }}{u_8} = 5.6284,{\text{ }}{v_8} = 6.2496\) A1
\(n = 8\;\;\;\)(must be an integer) A1 N2
[4 marks]
Total [14 marks]
Question
In an arithmetic sequence \({u_{10}} = 8,{\text{ }}{u_{11}} = 6.5\).
Write down the value of the common difference.
Find the first term.
Find the sum of the first 50 terms of the sequence.
Answer/Explanation
Markscheme
\(d = – 1.5\) A1 N1
[1 mark]
METHOD 1
valid approach (M1)
eg\(\;\;\;{u_{10}} = {u_1} + 9d,{\text{ }}8 = {u_1} – 9( – 1.5)\)
correct working (A1)
eg\(\;\;\;8 = {u_1} + 9d,{\text{ }}6.5 = {u_1} + 10d,{\text{ }}{u_1} = 8 – 9( – 1.5)\)
\({u_1} = 21.5\) A1 N2
METHOD 2
attempt to list 3 or more terms in either direction (M1)
eg\(\;\;\;9.5,{\text{ }}11,{\text{ }}12.5,{\text{ }} \ldots ;{\text{ }}5,{\text{ }}3.5,{\text{ }}2,{\text{ }} \ldots {\text{ }} \ldots \)
correct list of 4 or more terms in correct direction (A1)
eg\(\;\;\;9.5,{\text{ }}11,{\text{ }}12.5,{\text{ }}14\)
\({u_1} = 21.5\) A1 N2
[3 marks]
correct expression (A1)
eg\(\;\;\;\frac{{50}}{2}\left( {2(21.5) + 49( – 1.5)} \right),{\text{ }}\frac{{50}}{2}(21.5 – 52),{\text{ }}\sum\limits_{k = 1}^{50} {21.5 + (k – 1)( – 1.5)} \)
\({\text{sum}} = – 762.5\;\;\;{\text{(exact)}}\) A1 N2
[2 marks]
Total [6 marks]
Question
Ramiro walks to work each morning. During the first minute he walks \(80\) metres. In each subsequent minute he walks \(90\% \) of the distance walked during the previous minute.
The distance between his house and work is \(660\) metres. Ramiro leaves his house at 08:00 and has to be at work by 08:15.
Explain why he will not be at work on time.
Answer/Explanation
Markscheme
METHOD 1
recognize that the distance walked each minute is a geometric sequence (M1)
eg\(\;\;\;r = 0.9\), valid use of \(0.9\)
recognize that total distance walked is the sum of a geometric sequence (M1)
eg\(\;\;\;{S_n},{\text{ }}a\left( {\frac{{1 – {r^n}}}{{1 – r}}} \right)\)
correct substitution into the sum of a geometric sequence (A1)
eg\(\;\;\;80\left( {\frac{{1 – {{0.9}^n}}}{{1 – 0.9}}} \right)\)
any correct equation with sum of a geometric sequence (A1)
eg\(\;\;\;80\left( {\frac{{{{0.9}^n} – 1}}{{0.9 – 1}}} \right) = 660,{\text{ }}1 – {0.9^n} = \frac{{66}}{{80}}\)
attempt to solve their equation involving the sum of a GP (M1)
eg\(\;\;\;\)graph, algebraic approach
\(n = 16.54290788\) A1
since \(n > 15\) R1
he will be late AG N0
Note: Do not award the R mark without the preceding A mark.
METHOD 2
recognize that the distance walked each minute is a geometric sequence (M1)
eg\(\;\;\;r = 0.9\), valid use of \(0.9\)
recognize that total distance walked is the sum of a geometric sequence (M1)
eg\(\;\;\;{S_n},{\text{ }}a\left( {\frac{{1 – {r^n}}}{{1 – r}}} \right)\)
correct substitution into the sum of a geometric sequence (A1)
eg\(\;\;\;80\left( {\frac{{1 – {{0.9}^n}}}{{1 – 0.9}}} \right)\)
attempt to substitute \(n = 15\) into sum of a geometric sequence (M1)
eg\(\;\;\;{S_{15}}\)
correct substitution (A1)
eg\(\;\;\;80\left( {\frac{{{{0.9}^{15}} – 1}}{{0.9 – 1}}} \right)\)
\({S_{15}} = 635.287\) A1
since \(S < 660\) R1
he will not be there on time AG N0
Note: Do not award the R mark without the preceding A mark.
METHOD 3
recognize that the distance walked each minute is a geometric sequence (M1)
eg\(\;\;\;r = 0.9\), valid use of \(0.9\)
recognize that total distance walked is the sum of a geometric sequence (M1)
eg\(\;\;\;{S_n},{\text{ }}a\left( {\frac{{1 – {r^n}}}{{1 – r}}} \right)\)
listing at least 5 correct terms of the GP (M1)
15 correct terms A1
\(80,{\rm{ }}72,{\rm{ }}64.8,{\rm{ }}58.32,{\rm{ }}52.488,{\rm{ }}47.2392,{\rm{ }}42.5152,{\rm{ }}38.2637,{\rm{ }}34.4373,{\rm{ }}30.9936,{\rm{ }}27.8942,{\rm{ }}25.1048,{\rm{ }}22.59436,{\rm{ }}20.3349,{\rm{ }}18.3014\)
attempt to find the sum of the terms (M1)
eg\(\;\;\;{S_{15}},{\text{ }}80 + 72 + 64.8 + 58.32 + 52.488 + \ldots + 18.301433\)
\({S_{15}} = 635.287\) A1
since \(S < 660\) R1
he will not be there on time AG N0
Note: Do not award the R mark without the preceding A mark.
[7 marks]
Question
The first three terms of a geometric sequence are \({u_1} = 0.64,{\text{ }}{u_2} = 1.6\), and \({u_3} = 4\).
Find the value of \(r\).
Find the value of \({S_6}\).
Find the least value of \(n\) such that \({S_n} > 75\,000\).
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\;\;\;\frac{{{u_1}}}{{{u_2}}},{\text{ }}\frac{4}{{1.6}},{\text{ }}1.6 = r(0.64)\)
\(r = 2.5\;\;\;\left( { = \frac{5}{2}} \right)\) A1 N2
[2 marks]
correct substitution into \({S_6}\) (A1)
eg\(\;\;\;\frac{{0.64({{2.5}^6} – 1)}}{{2.5 – 1}}\)
\({S_6} = 103.74\) (exact), \(104\) A1 N2
[2 marks]
METHOD 1 (analytic)
valid approach (M1)
eg\(\;\;\;\frac{{0.64({{2.5}^n} – 1)}}{{2.5 – 1}} > 75\,000,{\text{ }}\frac{{0.64({{2.5}^n} – 1)}}{{2.5 – 1}} = 75\,000\)
correct inequality (accept equation) (A1)
eg\(\;\;\;n > 13.1803,{\text{ }}n = 13.2\)
\(n = 14\) A1 N1
METHOD 2 (table of values)
both crossover values A2
eg\(\;\;\;{S_{13}} = 63577.8,{\text{ }}{S_{14}} = 158945\)
\(n = 14\) A1 N1
[3 marks]
Total [7 marks]
Question
In a geometric sequence, the fourth term is 8 times the first term. The sum of the first 10 terms is 2557.5. Find the 10th term of this sequence.
Answer/Explanation
Markscheme
correct equation to find \(r\) (A1)
eg\(\,\,\,\,\,\)\({u_1}{r^3} = 8{u_1},{\text{ }}{r^3} = 8\)
\(r = 2\) (seen anywhere) (A1)
correct equation to find \({u_1}\) A1
eg\(\,\,\,\,\,\)\({u_1}({2^{10}} – 1) = 2557.5,{\text{ }}{u_1} = \frac{{2557.5}}{{{r^{10}} – 1}}(r – 1)\)
\({u_1} = 2.5\) (A1)
\({u_{10}} = 2.5{(2)^9}\) (M1)
1280 A1 N4
[6 marks]
Question
The first three terms of an arithmetic sequence are \({u_1} = 0.3,{\text{ }}{u_2} = 1.5,{\text{ }}{u_3} = 2.7\).
Find the common difference.
Find the 30th term of the sequence.
Find the sum of the first 30 terms.
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)\(1.5 – 0.3,{\text{ }}1.5 – 2.7,{\text{ }}2.7 = 0.3 + 2d\)
\(d = 1.2\) A1 N2
[2 marks]
correct substitution into term formula (A1)
eg\(\,\,\,\,\,\)\(0.3 + 1.2(30 – 1),{\text{ }}{u_{30}} = 0.3 + 29(1.2)\)
\({u_{30}} = 35.1\) A1 N2
[2 marks]
correct substitution into sum formula (A1)
eg\(\,\,\,\,\,\)\({S_{30}} = \frac{{30}}{2}(0.3 + 35.1),{\text{ }}\frac{{30}}{2}\left( {2(0.3) + 29(1.2)} \right)\)
\({S_{30}} = 531\) A1 N2
[2 marks]
Question
Consider a geometric sequence where the first term is 768 and the second term is 576.
Find the least value of \(n\) such that the \(n\)th term of the sequence is less than 7.
Answer/Explanation
Markscheme
attempt to find \(r\) (M1)
eg\(\,\,\,\,\,\)\(\frac{{576}}{{768}},{\text{ }}\frac{{768}}{{576}},{\text{ }}0.75\)
correct expression for \({u_n}\) (A1)
eg\(\,\,\,\,\,\)\(768{(0.75)^{n – 1}}\)
EITHER (solving inequality)
valid approach (accept equation) (M1)
eg\(\,\,\,\,\,\)\({u_n} < 7\)
valid approach to find \(n\) M1
eg\(\,\,\,\,\,\)\(768{(0.75)^{n – 1}} = 7,{\text{ }}n – 1 > {\log _{0.75}}\left( {\frac{7}{{768}}} \right)\), sketch
correct value
eg\(\,\,\,\,\,\)\(n = 17.3301\) (A1)
\(n = 18\) (must be an integer) A1 N2
OR (table of values)
valid approach (M1)
eg\(\,\,\,\,\,\)\({u_n} > 7\), one correct crossover value
both crossover values, \({u_{17}} = 7.69735\) and \({u_{18}} = 5.77301\) A2
\(n = 18\) (must be an integer) A1 N2
OR (sketch of functions)
valid approach M1
eg\(\,\,\,\,\,\)sketch of appropriate functions
valid approach (M1)
eg\(\,\,\,\,\,\)finding intersections or roots (depending on function sketched)
correct value
eg\(\,\,\,\,\,\)\(n = 17.3301\) (A1)
\(n = 18\) (must be an integer) A1 N2
[6 marks]
Question
Let \(f\left( x \right) = {{\text{e}}^{2\,{\text{sin}}\left( {\frac{{\pi x}}{2}} \right)}}\), for x > 0.
The k th maximum point on the graph of f has x-coordinate xk where \(k \in {\mathbb{Z}^ + }\).
Given that xk + 1 = xk + a, find a.
Hence find the value of n such that \(\sum\limits_{k = 1}^n {{x_k} = 861} \).
Answer/Explanation
Markscheme
valid approach to find maxima (M1)
eg one correct value of xk, sketch of f
any two correct consecutive values of xk (A1)(A1)
eg x1 = 1, x2 = 5
a = 4 A1 N3
[4 marks]
recognizing the sequence x1, x2, x3, …, xn is arithmetic (M1)
eg d = 4
correct expression for sum (A1)
eg \(\frac{n}{2}\left( {2\left( 1 \right) + 4\left( {n – 1} \right)} \right)\)
valid attempt to solve for n (M1)
eg graph, 2n2 − n − 861 = 0
n = 21 A1 N2
[4 marks]
Question
The first term of an infinite geometric sequence is 4. The sum of the infinite sequence is 200.
Find the common ratio.
Find the sum of the first 8 terms.
Find the least value of n for which Sn > 163.
Answer/Explanation
Markscheme
correct substitution into infinite sum (A1)
eg \(200 = \frac{4}{{1 – r}}\)
r = 0.98 (exact) A1 N2
[2 marks]
correct substitution (A1)
\(\frac{{4\left( {1 – {{0.98}^8}} \right)}}{{1 – 0.98}}\)
29.8473
29.8 A1 N2
[2 marks]
attempt to set up inequality (accept equation) (M1)
eg \(\frac{{4\left( {1 – {{0.98}^n}} \right)}}{{1 – 0.98}} > 163,\,\,\frac{{4\left( {1 – {{0.98}^n}} \right)}}{{1 – 0.98}} = 163\)
correct inequality for n (accept equation) or crossover values (A1)
eg n > 83.5234, n = 83.5234, S83 = 162.606 and S84 = 163.354
n = 84 A1 N1
[3 marks]