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Date	November 2017	Marks available	5	Reference code	17N.1.SL.TZ0.S_10
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Show that	Question number	S_10	Adapted from	N/A

Question

The following diagram shows [AB], with length 2 cm. The line is divided into an infinite number of line segments. The diagram shows the first three segments.

N17/5/MATME/SP1/ENG/TZ0/10.a

The length of the line segments are $p{\text{ cm}},{\text{ }}{p^2}{\text{ cm}},{\text{ }}{p^3}{\text{ cm}},{\text{ }} \ldots$ , where $0 < p < 1$ .

Show that $p = \frac{2}{3}$ .

[5]

The following diagram shows [CD], with length $b{\text{ cm}}$ , where $b > 1$ . Squares with side lengths $k{\text{ cm}},{\text{ }}{k^2}{\text{ cm}},{\text{ }}{k^3}{\text{ cm}},{\text{ }} \ldots$ , where $0 < k < 1$ , are drawn along [CD]. This process is carried on indefinitely. The diagram shows the first three squares.

N17/5/MATME/SP1/ENG/TZ0/10.b

The total sum of the areas of all the squares is $\frac{9}{{16}}$ . Find the value of $b$ .

[9]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

infinite sum of segments is 2 (seen anywhere) (A1)

eg $\,\,\,\,\,$ $p + {p^2} + {p^3} + \ldots = 2,{\text{ }}\frac{{{u_1}}}{{1 - r}} = 2$

recognizing GP (M1)

eg $\,\,\,\,\,$ ratio is $p,{\text{ }}\frac{{{u_1}}}{{1 - r}},{\text{ }}{u_n} = {u_1} \times {r^{n - 1}},{\text{ }}\frac{{{u_1}({r^n} - 1)}}{{r - 1}}$

correct substitution into ${S_\infty }$ formula (may be seen in equation) A1

eg $\,\,\,\,\,$ $\frac{p}{{1 - p}}$

correct equation (A1)

eg $\,\,\,\,\,$ $\frac{p}{{1 - p}} = 2,{\text{ }}p = 2 - 2p$

correct working leading to answer A1

eg $\,\,\,\,\,$ $3p = 2,{\text{ }}2 - 3p = 0$

$p = \frac{2}{3}{\text{ (cm)}}$ AG N0

[5 marks]

recognizing infinite geometric series with squares (M1)

eg $\,\,\,\,\,$ ${k^2} + {k^4} + {k^6} + \ldots ,{\text{ }}\frac{{{k^2}}}{{1 - {k^2}}}$

correct substitution into ${S_\infty } = \frac{9}{{16}}$ (must substitute into formula) (A2)

eg $\,\,\,\,\,$ $\frac{{{k^2}}}{{1 - {k^2}}} = \frac{9}{{16}}$

correct working (A1)

eg $\,\,\,\,\,$ $16{k^2} = 9 - 9{k^2},{\text{ }}25{k^2} = 9,{\text{ }}{k^2} = \frac{9}{{25}}$

$k = \frac{3}{5}$ (seen anywhere) A1

valid approach with segments and CD (may be seen earlier) (M1)

eg $\,\,\,\,\,$ $r = k,{\text{ }}{S_\infty } = b$

correct expression for $b$ in terms of $k$ (may be seen earlier) (A1)

eg $\,\,\,\,\,$ $b = \frac{k}{{1 - k}},{\text{ }}b = \sum\limits_{n = 1}^\infty {{k^n},{\text{ }}b = k + {k^2} + {k^3} + \ldots }$

substituting their value of $k$ into their formula for $b$ (M1)

eg $\,\,\,\,\,$ $\frac{{\frac{3}{5}}}{{1 - \frac{3}{5}}},{\text{ }}\frac{{\left( {\frac{3}{5}} \right)}}{{\left( {\frac{2}{5}} \right)}}$

$b = \frac{3}{2}$ A1 N3

[9 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.3—Geometric sequences and series

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