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.

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

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.

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

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]