Math Repository
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topic
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2007
P1 Q10
Topical
AP/GP
07 P1 Q10
2007 H2 Mathematics Paper 1 Question 10
Arithmetic and Geometric Progressions (APs, GPs)
Answers
(ii)
The geometric series is convergent since
−
1
<
r
=
2
3
<
1.
{-1 < r = \frac{2}{3} < 1.}
−
1
<
r
=
3
2
<
1.
S
∞
=
3
a
{S_{\infty} = 3a}
S
∞
=
3
a
(iii)
{
n
∈
Z
:
6
≤
n
≤
13
}
{\{ n \in \mathbb{Z} : 6 \leq n \leq 13 \}}
{
n
∈
Z
:
6
≤
n
≤
13
}
Full solutions
(i)
a
=
a
a
+
3
d
=
a
r
a
+
5
d
=
a
r
2
\begin{align} &&\quad a &= a \\ &&\quad a+3d &= ar \\ &&\quad a+5d &= ar^{2} \\ \end{align}
a
a
+
3
d
a
+
5
d
=
a
=
a
r
=
a
r
2
Taking
(
2
)
−
(
1
)
{(2)-(1)}
(
2
)
−
(
1
)
,
3
d
=
a
r
−
a
\begin{equation}\qquad 3d = ar-a \end{equation}
3
d
=
a
r
−
a
Taking
(
3
)
−
(
1
)
{(3)-(1)}
(
3
)
−
(
1
)
,
5
d
=
a
r
2
−
a
\begin{equation}\qquad 5d = ar^{2}-a \end{equation}
5
d
=
a
r
2
−
a
From
(
4
)
{(4)}
(
4
)
and
(
5
)
{(5)}
(
5
)
,
a
r
−
a
3
=
a
r
2
−
a
5
5
a
r
−
5
a
=
3
b
r
2
−
3
a
3
a
r
2
−
5
a
+
2
a
=
0
3
r
2
−
5
r
+
2
=
0
■
\begin{gather*} \frac{ar-a}{3} = \frac{ar^{2}-a}{5} \\ 5ar-5a = 3br^{2}-3a \\ 3ar^2 -5a + 2a = 0 \\ 3 r^2 - 5 r + 2 = 0 \; \blacksquare \end{gather*}
3
a
r
−
a
=
5
a
r
2
−
a
5
a
r
−
5
a
=
3
b
r
2
−
3
a
3
a
r
2
−
5
a
+
2
a
=
0
3
r
2
−
5
r
+
2
=
0
■
(ii)
(
3
r
−
2
)
(
r
−
1
)
=
0
r
=
2
3
or
r
=
1
\begin{gather*} (3 r - 2)(r - 1) = 0 \\ r = \frac{2}{3} \; \textrm{ or } \; r = 1 \end{gather*}
(
3
r
−
2
)
(
r
−
1
)
=
0
r
=
3
2
or
r
=
1
r
=
1
{r=1}
r
=
1
is rejected since
d
{d}
d
is non-zero
Hence
r
=
2
3
{r=\frac{2}{3}}
r
=
3
2
Since
−
1
<
r
=
2
3
<
1
,
{-1 < r = \frac{2}{3} < 1,}
−
1
<
r
=
3
2
<
1
,
the geometric series is convergent
■
{\blacksquare}
■
S
∞
=
a
1
−
r
=
a
1
−
2
3
=
3
a
■
\begin{align*} S_{\infty} &= \frac{a}{1-r} \\ &= \frac{a}{1-\frac{2}{3}} \\ &= 3a \; \blacksquare \end{align*}
S
∞
=
1
−
r
a
=
1
−
3
2
a
=
3
a
■
(iii)
d
=
a
r
−
a
3
=
a
(
2
3
)
−
a
3
=
−
1
9
a
\begin{align*} d &= \frac{ar-a}{3} \\ &= \frac{a\left(\frac{2}{3}\right)-a}{3} \\ &= - \frac{1}{9}a \end{align*}
d
=
3
a
r
−
a
=
3
a
(
3
2
)
−
a
=
−
9
1
a
S
>
4
a
n
2
(
2
a
+
(
n
−
1
)
d
)
>
4
a
n
2
(
2
a
+
(
n
−
1
)
(
−
1
9
a
)
)
>
4
a
18
a
n
−
a
n
2
+
a
n
>
72
a
\begin{gather*} S > 4a \\ \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) > 4a \\ \frac{n}{2}\Big(2a + (n-1)\left(- \frac{1}{9}a\right) \Big) > 4a \\ 18an - an^2 + an > 72a \\ \end{gather*}
S
>
4
a
2
n
(
2
a
+
(
n
−
1
)
d
)
>
4
a
2
n
(
2
a
+
(
n
−
1
)
(
−
9
1
a
)
)
>
4
a
18
an
−
a
n
2
+
an
>
72
a
Since
a
>
0
,
{a>0,}
a
>
0
,
n
2
−
19
n
+
72
<
0
5.2280
<
n
<
13.772
\begin{gather*} n^2 - 19 n + 72 < 0 \\ 5.2280 < n < 13.772 \end{gather*}
n
2
−
19
n
+
72
<
0
5.2280
<
n
<
13.772
Set of possible values of
n
:
{n:}
n
:
{
n
∈
Z
:
6
≤
n
≤
13
}
■
\{ n \in \mathbb{Z} : 6 \leq n \leq 13 \} \; \blacksquare
{
n
∈
Z
:
6
≤
n
≤
13
}
■
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