2017 H2 Mathematics Paper 2 Question 2

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(i)

d=32{d=\frac{3}{2}}

(ii)

r=−1.45{r=-1.45} or r=1.21{r=1.21}

(iii)

Smallest n=42{\textrm{Smallest } n = 42}

Full solutions

(i)

S13=156n2(2a+(n−1)d)=156132(2(3)+(13−1)d)=1566+12d=24d=32  ■\begin{gather*} S_{13} = 156 \\ \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) = 156 \\ \frac{13}{2}\Big( 2(3) + (13-1)d \Big) = 156 \\ 6 + 12d = 24 \\ d = \frac{3}{2} \; \blacksquare \end{gather*}

(ii)

S13=156a(1−rn)1−r=1563(1−r13)1−r=1561−r13=52(1−r)r13−52r+51=0  ■\begin{gather*} S_{13} = 156 \\ \frac{a\left(1-r^{n}\right)}{1-r} = 156 \\ \frac{3(1-r^{13})}{1-r} = 156 \\ 1-r^{13} = 52(1-r) \\ r^{13} - 52 r + 51 = 0 \; \blacksquare \end{gather*}
r=1{r=1} is a root of the equation because
113−52(1)+51=1−52+51=0\begin{align*} & 1^{13}-52(1)+51 \\ &= 1-52+51 \\ &= 0 \end{align*}
However, if r=1,{r=1, } then the geometric progression is a constant sequence
3,3,3,…3, 3, 3, \ldots
Then the sum of the first 13{13} terms is 13×3=39≠156{13\times 3 = 39 \neq 156}
Hence the common ratio cannot be 1  ■{1 \; \blacksquare}
Using a GC,
r=−1.45   or   r=1.21  ■r=-1.45 \; \textrm{ or } \; r=1.21 \; \blacksquare

(iii)

r=1.2100{r=1.2100}
arn−1>100(a+(n−1)d)3(1.21)n−1−150n−150>0\begin{gather*} ar^{n-1} > 100 \Big( a + (n-1)d \Big) \\ 3(1.21)^{n-1} - 150n - 150 > 0 \end{gather*}
Using a GC,
Smallest n=42  ■\textrm{Smallest } n = 42 \; \blacksquare