IB DP Maths Topic 1.1 Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series SL Paper 2

 

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Question

Consider the infinite geometric sequence \(3000{\text{, }}- 1800{\text{, }}1080{\text{, }} – 648, \ldots \) .

Find the common ratio.

[2]
a.

Find the 10th term.

[2]
b.

Find the exact sum of the infinite sequence.

[2]
c.
Answer/Explanation

Markscheme

evidence of dividing two terms     (M1)

e.g. \( – \frac{{1800}}{{3000}}\) , \( – \frac{{1800}}{{1080}}\)

\(r = – 0.6\)     A1     N2

[2 marks]

a.

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]

b.

evidence of substituting into the formula for the infinite sum     (M1)

e.g. \(S = \frac{{3000}}{{1.6}}\)

\(S = 1875\)     A1     N2

[2 marks]

c.

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.

[6]
a(i) and (ii).

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.

[6]
b(i) and (ii).

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?

[5]
c(i) and (ii).
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]

a(i) and (ii).

(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]

b(i) and (ii).

(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]

c(i) and (ii).

Question

In an arithmetic series, the first term is –7 and the sum of the first 20 terms is 620.

Find the common difference.

[3]
a.

Find the value of the 78th term.

[2]
b.
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]

a.

correct substitution \({u_{78}} = – 7 + 77(4)\)     (A1)

= 301     A1     N2

[2 marks]

b.

Question

In a geometric series, \({u_1} = \frac{1}{{81}}\) and \({u_4} = \frac{1}{3}\) .

Find the value of \(r\) .

[3]
a.

Find the smallest value of n for which \({S_n} > 40\) .

[4]
b.
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]

a.

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]

b.

Question

Expand \(\sum\limits_{r = 4}^7 {{2^r}} \) as the sum of four terms.

[1]
a.

(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.

[6]
b.
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]

a.

(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]

b.

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.

[1]
a.

Find the number of terms in the sequence.

[3]
b.

Find the sum of the sequence.

[2]
c.
Answer/Explanation

Markscheme

common difference is 6     A1     N1

[1 mark]

a.

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]

b.

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]

c.

Question

An arithmetic sequence, \({u_1}{\text{, }}{u_2}{\text{, }}{u_3} \ldots ,\) has \(d = 11\) and \({u_{27}} = 263\) .

Find \({u_1}\).

[2]
a.

(i)     Given that \({u_n} = 516\) , find the value of n .

(ii)    For this value of n , find \({S_n}\) .

[4]
b(i) and (ii).
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]

a.

(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]

b(i) and (ii).

Question

The nth term of an arithmetic sequence is given by \({u_n} = 5 + 2n\) .

Write down the common difference.

[1]
a.

(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.

[5]
b(i) and (ii).
Answer/Explanation

Markscheme

\(d = 2\)     A1     N1

[1 mark]

a.

(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]

b(i) and (ii).

Question

In an arithmetic sequence \({u_1} = 7\) , \({u_{20}} = 64\) and \({u_n} = 3709\) .

Find the value of the common difference.

[3]
a.

Find the value of n .

[2]
b.
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]

a.

correct substitution into formula for \({u_n}\)     A1

e.g. \(3709 = 7 + 3(n – 1)\) , \(3709 = 3n + 4\)

\(n = 1235\)     A1     N1

[2 marks]

b.

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.

[4]
a(i) and (ii).

Consider an arithmetic sequence with n terms, with first term (\( – 36\)) and eighth term (\( – 8\)) .

(i)     Find the common difference.

(ii)    Show that \({S_n} = 2{n^2} – 38n\) .

[5]
b(i) and (ii).

The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n .

[5]
c.
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]

a(i) and (ii).

(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]

b(i) and (ii).

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]

c.

Question

The first three terms of an arithmetic sequence are 36, 40, 44,….

(i)     Write down the value of d .

(ii)    Find \({u_8}\) .

[3]
a(i) and (ii).

(i)     Show that \({S_n} = 2{n^2} + 34n\) .

(ii)    Hence, write down the value of \({S_{14}}\) .

[3]
b(i) and (ii).
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]

a(i) and (ii).

(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]

b(i) and (ii).

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.

[4]
a.

The first term of a geometric sequence is 200 and the sum of the first four terms is 324.8.

Find the tenth term.

[2]
b.
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]

a.

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]

b.

Question

The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .

Find the common difference.

[2]
a.

The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .

Find the 28th term of the sequence.

[2]
b.

The first three terms of an arithmetic sequence are \(5\) , \(6.7\) , \(8.4\) .

Find the sum of the first 28 terms.

[2]
c.
Answer/Explanation

Markscheme

valid method     (M1)

e.g. subtracting terms, using sequence formula

\(d = 1.7\)     A1     N2

[2 marks]

a.

correct substitution into term formula     (A1)

e.g. \(5 + 27(1.7)\)

28th term is 50.9 (exact)     A1     N2

[2 marks]

b.

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]

c.

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 \(n\) .

[7]
.

Write down the value of \(d\) .

[1]
a.

Find

  (i)      \({u_{100}}\) ;

  (ii)      \({S_{100}}\) .

[4]
b.

Given that \({u_n} = 1502\) , find the value of \(n\) .

[2]
c.
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]

a.

(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]

b.

correct substitution into term formula     (A1)

eg     \(1502 = 5 + 3(n – 1)\) ,  \(1502 = 3n + 2\)

\(n = 500\)     A1     N2

[2 marks]

Total [7 marks]

c.

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 \({u_5}\).

[5]
a.

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 \(k\).

[5]
b.

Find the smallest value of \(n\) for which \({v_n} > {u_n}\).

[5]
c.
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]

a.

(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]

b.

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]

c.

Question

In an arithmetic sequence \({u_{10}} = 8,{\text{ }}{u_{11}} = 6.5\).

Write down the value of the common difference.

[1]
a.

Find the first term.

[3]
b.

Find the sum of the first 50 terms of the sequence.

[2]
c.
Answer/Explanation

Markscheme

\(d =  – 1.5\)     A1     N1

[1 mark]

a.

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]

b.

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]

c.

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 \(r\).

[2]
a.

Find the value of \({S_6}\).

[2]
b.

Find the least value of \(n\) such that \({S_n} > 75\,000\).

[3]
c.
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]

a.

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]

b.

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]

c.

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.

[2]
a.

Find the 30th term of the sequence.

[2]
b.

Find the sum of the first 30 terms.

[2]
c.
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]

a.

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]

b.

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]

c.

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.

[4]
a.

Hence find the value of n such that \(\sum\limits_{k = 1}^n {{x_k} = 861} \).

[4]
b.
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]

a.

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, 2n2n − 861 = 0

n = 21       A1 N2

[4 marks]

b.

Question

The first term of an infinite geometric sequence is 4. The sum of the infinite sequence is 200.

Find the common ratio.

[2]
a.

Find the sum of the first 8 terms.

[2]
b.

Find the least value of n for which Sn > 163.

[3]
c.
Answer/Explanation

Markscheme

correct substitution into infinite sum      (A1)
eg  \(200 = \frac{4}{{1 – r}}\)

= 0.98 (exact)     A1 N2

[2 marks]

a.

correct substitution     (A1)

\(\frac{{4\left( {1 – {{0.98}^8}} \right)}}{{1 – 0.98}}\)

29.8473

29.8    A1 N2

[2 marks]

b.

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]

c.
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