Sec 1 Maths: Number Patterns & Sequences — practice questions & worked solutions

(i) Figure 4 is a $4\times 4$ grid of squares: a $5\times 5$ array of circles joined by sticks.

(ii) Figure 4: $40$ sticks, $25$ circles. Figure 5: $60$ sticks, $36$ circles.

Number of sticks in Figure $n$ is $2n(n+1)$ and number of circles is $(n+1)^{2}$. Figure 4 sticks $=2(4)(5)=40$, circles $=5^{2}=25$. Figure 5 sticks $=2(5)(6)=60$, circles $=6^{2}=36$.

(iii) Set sticks $=842$:

$$\begin{aligned}2n(n+1)&=842\\ n(n+1)&=421\end{aligned}$$

Since $20\times 21=420$ and $21\times 22=462$, there is no whole number $n$ with $n(n+1)=421$. Hence it is not possible.

(iv)(a) Number of circles in Figure $n$ is $(n+1)^{2}$.

(b) Set circles $=2500$:

$$\begin{aligned}(n+1)^{2}&=2500\\ n+1&=50\\ n&=49\end{aligned}$$

The largest Figure is Figure $49$.

(a)

Arithmetic sequence, first term $3$, common difference $2$.

$$\begin{aligned}T_n &= 3 + 2(n – 1) = 2n + 1\end{aligned}$$

(b)(i)

The pattern is $n^{2} + (2n + 1) = (n + 1)^{2}$. Row $10$:

$$\begin{aligned}10^{2} + 21 &= 121 = 11^{2}\end{aligned}$$

Row $n$:

$$\begin{aligned}n^{2} + 2n + 1 &= (n + 1)^{2}\end{aligned}$$

(b)(ii)

The sequence here is the squares $(n + 1)^{2}$ (i.e. $4, 9, 16, 25, \ldots$). For $216$ to be a term it must be a perfect square; $\sqrt{216} \approx 14.7$ is not a whole number, so $216$ is not a term.

(b)(iii)

Consecutive squares $(n + 1)^{2}$ and $(n + 2)^{2}$ differ by

$$\begin{aligned}(n + 2)^{2} – (n + 1)^{2} &= 2n + 3 = 97\end{aligned}$$
$$\begin{aligned}2n &= 94\end{aligned}$$
$$\begin{aligned}n &= 47\end{aligned}$$
$$\begin{aligned}(n + 1)^{2} &= 48^{2} = 2304\end{aligned}$$
$$\begin{aligned}(n + 2)^{2} &= 49^{2} = 2401\end{aligned}$$

The two consecutive terms are $2304$ and $2401$.

The distances increase by $0.4$ km each day (arithmetic sequence with first term $1$). Day 5:

$$\begin{aligned}1+(4)(0.4)&=2.6\end{aligned}$$

Day $n$:

$$\begin{aligned}1+(n-1)(0.4)&=0.4n+0.6\end{aligned}$$

The distances double each day (geometric, first term $0.5$, ratio $2$):

$$\begin{aligned}0.5\times2^{n-1}&=\frac{1}{2}\times2^{n-1}\end{aligned}$$
$$\begin{aligned}&=2^{n-2}\text{ km}\end{aligned}$$

Require $2^{n-2}\ge 21$. Testing powers of $2$:

$$\begin{aligned}2^4&=16<21\end{aligned}$$
$$\begin{aligned}2^5&=32\ge 21\end{aligned}$$

So $n-2=5$, giving $n=7$.

(i)

The sequence is arithmetic with first term 1 and common difference 3.

$$\begin{aligned}T_n &= 1 + (n – 1)(3) \\ &= 3n – 2\end{aligned}$$

(ii)

Writing the terms of $Q_n$ with numerators $1, 4, 7, 10, 13, \ldots = T_n = 3n – 2$ and denominators $2, 4, 6, 8, 10, \ldots = U_n = 2n$:

$$\begin{aligned}Q_n &= \frac{T_n}{U_n} \\ &= \frac{3n – 2}{2n}\end{aligned}$$

(iii)

Since $Q_n$ is increasing, $Q_{2k} > Q_k$.

$$\begin{aligned}Q_{2k} – Q_k &= \frac{1}{44} \\ \frac{3(2k) – 2}{2(2k)} – \frac{3k – 2}{2k} &= \frac{1}{44} \\ \frac{6k – 2}{4k} – \frac{6k – 4}{4k} &= \frac{1}{44} \\ \frac{2}{4k} &= \frac{1}{44} \\ \frac{1}{2k} &= \frac{1}{44} \\ 2k &= 44 \\ k &= 22\end{aligned}$$

(i)

The pattern for line $n$:

$$\begin{aligned}5^2 – 1^2 &= (4)(6)\\ 6^2 – 1^2 &= (5)(7)\\ n^2 – 1^2 &= (n-1)(n+1)\end{aligned}$$

(ii)

For the second pattern:

$$\begin{aligned}n^2 – 2^2 &= (n-2)(n+2)\end{aligned}$$

(iii)

In general $a^2 – b^2 = (a-b)(a+b)$, so:

$$\begin{aligned}55^2 – 45^2 &= (55-45)(55+45)\\ &= (10)(100)\\ &= 1000\end{aligned}$$

(a)

Common difference $= 26 – 33 = -7$.

$$\begin{aligned}T_7 &= 12 – 7 – 7 – 7 \\ T_7 &= -9\end{aligned}$$

(b)

$$\begin{aligned}T_n &= 33 + (n-1)(-7) \\ T_n &= 40 – 7n\end{aligned}$$

(c)

Set $40 – 7n = -206$:

$$\begin{aligned}7n &= 246 \\ n &= 35\tfrac{1}{7}\end{aligned}$$

Since $n$ is not a whole number, $-206$ is not in the sequence.

(a)

$$\begin{aligned}T_3 &= \frac{3(3) + 2}{47 – 2(3)} \\ &= \frac{11}{41}\end{aligned}$$

(b)

$$\begin{aligned}\frac{3k + 2}{47 – 2k} &= 13 \\ 3k + 2 &= 13(47 – 2k) \\ 3k + 2 &= 611 – 26k \\ 29k &= 609 \\ k &= 21\end{aligned}$$

(c)(i)

$n$ is a positive integer, so $n > 0$, hence $3n > 0$ and therefore $3n + 2 > 0$ for all such $n$.

(c)(ii)

Since $3n + 2 > 0$, $T_n < 0$ requires $47 – 2n < 0$, i.e. $n > 23.5$.

$$\begin{aligned}n &= 24\end{aligned}$$

(a)

The numerators are $1,2,3,4,\ldots$ and the denominators are $(n+1)^{2}$, i.e. $4,9,16,25,36,49,\ldots$ Next two terms:

$$\frac{5}{36},\ \frac{6}{49}$$

(b)

First term $8$, common difference $-3$:

$$\begin{aligned}T_n &= 8+(n-1)(-3)\\ &= -3n+11\end{aligned}$$

(a)

Common difference $=-7$, first term $=24$.

$$\begin{aligned}T_n &= 24+(n-1)(-7) \\ &= 31-7n\end{aligned}$$

(b)

Set $T_n=-700$:

$$\begin{aligned}31-7n &= -700 \\ 7n &= 731 \\ n &= 104.43\ldots\end{aligned}$$

Since $n$ is not a positive integer, $-700$ is not a term of the sequence.

(a)

Common difference $=7$, first term $=-20$:

$$T_n = -20 + (n-1)(7)$$
$$T_n = 7n – 27$$

(b)

Set $7n-27=100$:

$$7n = 127$$
$$n = 18\tfrac{1}{7}$$

Since $n$ is not a positive integer, $100$ is not a term in the sequence.