2019 H2 Mathematics Paper 1 Question 8

Arithmetic and Geometric Progressions (APs, GPs)

Answers

{k=13}

{r=-1, f\in\mathbb{R}}

{S_n = \begin{cases}f \quad &\textrm{ if } n \textrm{ odd} \\ 0 \quad &\textrm{ if } n \textrm{ even} \end{cases}}

{u_{11} = 63}

\begin{gather*} u_{k,\textrm{GP}} = S_{64,\textrm{AP}} \\ ar^{k-1} = \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) \\ a 2^{k-1} = \frac{64}{2} \left( 2a + 63(2a) \right) \\ 2^{k-1} = 4096 \\ 2^{k-1} = 2^{12} \\ k = 13 \; \blacksquare \end{gather*}

\begin{gather*} f + fr + fr^2 + fr^3 = 0 \\ f(1+r+r^2+r^3) = 0 \end{gather*}

Since

{f\neq 0,}

\begin{gather*} 1+r+r^2+r^3 = 0 \\ r = -1 \; \blacksquare \end{gather*}

{f}

can take any real value (except

{0,}

) as specified in the question.

f \in \mathbb{R}, f \neq 0 \; \blacksquare

Hence the sequence is

f, -f, f, -f, \ldots

S_n = \begin{cases}f \quad &\textrm{ if } n \textrm{ odd} \\ 0 \quad &\textrm{ if } n \textrm{ even} \end{cases} \; \blacksquare

\begin{gather*} a + (a+d) + (a+2d) + (a+3d) = 14 \\ 4a + 6d = 14 \end{gather*}

\begin{equation} \qquad 2a + 3d = 7 \end{equation}

\begin{equation} \qquad a(a+d)(a+2d)(a+3d) = 0 \end{equation}

From

{(2), }

{a=0}

(NA since

{a}

is negative)

\begin{align} &&\; \textrm{ or} \qquad a&=-d \\ &&\; \textrm{ or} \qquad a&=-2d \\ &&\; \textrm{ or} \qquad a&=-3d \\ \end{align}

Solving

{(1)}

and

{(4),}

a=14, \; d=- 7

Solving

{(1)}

and

{(5),}

a=7, \; d=- \frac{7}{3}

Both these cases are rejected since

{a}

is negative

Solving

{(1)}

and

{(3),}

a=- 7, \; d=7

\begin{align*} u_{11} &= a + \left( n - 1 \right) d \\ &= - 7 + \left( 11 - 1 \right) 7 \\ &= 63 \; \blacksquare \end{align*}