Consider the quadratic number pattern: 3 ; 7 ; 12 ; ... 3.1.1 Show that the general term of this number pattern is given by $T_n = \frac{1}{2}n^2 + \frac{5}{2}n$. 3.1.2 What number must be added to $T_n$ so that $T_n = 13 527$? 3.2 Given an arithmetic sequence with $T_i = 8$ and $T_2 = 11$. 3.2.1 Calculate the value of $n$ if $T_n = 41$. 3.2.2 A new arithmetic sequence $P$ is formed using the term position and the term value of the given arithmetic sequence. For the new sequence, $P_5 = 1$, $P_1 = 2$ and so forth. (a) Write down the value of $P_4$. (b) Calculate the value of the first term of the new arithmetic sequence.

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Consider the quadratic number pattern: 3 ; 7 ; 12 ; .. - NSC Mathematics - Question 3 - 2024 - Paper 1

Question 3

Consider the quadratic number pattern: 3 ; 7 ; 12 ; ... 3.1.1 Show that the general term of this number pattern is given by $T_n = \frac{1}{2}n^2 + \frac{5}{2}n$. ... show full transcript

Worked Solution & Example Answer:Consider the quadratic number pattern: 3 ; 7 ; 12 ; .. - NSC Mathematics - Question 3 - 2024 - Paper 1

Step 1

3.1.1 Show that the general term of this number pattern is given by $T_n = \frac{1}{2}n^2 + \frac{5}{2}n$

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Answer

To show that the general term is $T_n = \frac{1}{2}n^2 + \frac{5}{2}n$ , we can analyze the pattern by finding the first and second differences of the terms:

Given the sequence: 3, 7, 12
First differences: 4 (7-3), 5 (12-7)
Second differences: 1 (5-4)

We see that the second differences are constant, which suggests a quadratic formula of the form $T_n = an^2 + bn + c$ . Using the known terms:

$T_1 = 3 \Rightarrow a(1)^2 + b(1) + c = 3$
$T_2 = 7 \Rightarrow a(2)^2 + b(2) + c = 7$
$T_3 = 12 \Rightarrow a(3)^2 + b(3) + c = 12$

From these equations, we can derive values for $a$ , $b$ , and $c$ . Solving this gives:

$a = \frac{1}{2}, b = 5/2$ and clearly $c = 0$ . Thus, $T_n = \frac{1}{2}n^2 + \frac{5}{2}n$ .

Step 2

3.1.2 What number must be added to $T_n$ so that $T_n = 13 527$?

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Answer

Given that $T_n = \frac{1}{2}n^2 + \frac{5}{2}n$ and we want $T_n = 13 527$ ,

Set up the equation:

$\frac{1}{2}n^2 + \frac{5}{2}n = 13 527$

This simplifies to:

$n^2 + 5n - 27 054 = 0$

Using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4(1)(-27054)}}{2(1)}$

Calculating gives:

The roots will yield an $n$ value of either 161 or -167. Since $n$ must be positive, we take $n = 161$ .

Now, calculate $T_{161}$ :

$T_{161} = \frac{1}{2}(161)^2 + \frac{5}{2}(161) = 13 363$ .

Therefore, the number to be added is: $13 527 - 13 363 = 164$ .

Step 3

3.2.1 Calculate the value of $n$ if $T_n = 41$.

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Answer

Using the equation for the arithmetic sequence, $T_n = a + (n-1)d$ We know $T_1 = 8 \Rightarrow a = 8$ , and $T_2 = 11 \Rightarrow d = 3$ . Thus, we have: $T_n = 8 + (n - 1)3$ For $T_n = 41$ : $41 = 8 + (n - 1)3$ Solving for $n$ :

$41 - 8 = (n - 1)3 \Rightarrow 33 = (n - 1)3 \Rightarrow n - 1 = 11 \Rightarrow n = 12.$

Step 4

3.2.2(a) Write down the value of $P_4$.

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Answer

To find $P_4$ , we substitute into $P_n = a + (n-1)d$ . We established earlier:

$P_1 = 2$ , hence $a = 2$ and $d = \frac{1}{3}$ from previous calculations. Thus: $$P_4 = a + (4 - 1)d = 2 + 3\left(\frac{1}{3}\right) = 2 + 1 = 3.$

Step 5

3.2.2(b) Calculate the value of the first term of the new arithmetic sequence.

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Answer

From the relation of the arithmetic sequence,

We found $a = 2$ and $d = \frac{1}{3}$ , hence the first term: $P_1 = a = 2.$ Thus, the first term of the new arithmetic sequence is 2.

Consider the quadratic number pattern: 3 ; 7 ; 12 ; .. - NSC Mathematics - Question 3 - 2024 - Paper 1

Worked Solution & Example Answer:Consider the quadratic number pattern: 3 ; 7 ; 12 ; .. - NSC Mathematics - Question 3 - 2024 - Paper 1

3.1.1 Show that the general term of this number pattern is given by $T_n = \frac{1}{2}n^2 + \frac{5}{2}n$

3.1.2 What number must be added to $T_n$ so that $T_n = 13 527$?

3.2.1 Calculate the value of $n$ if $T_n = 41$.

3.2.2(a) Write down the value of $P_4$.

3.2.2(b) Calculate the value of the first term of the new arithmetic sequence.

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