Calculate the value of x

The sequence 4; 9; The pattern 4; 9; x; 37; …. is a quadratic pattern 1.5.1 Calculate the value of x. (3) 1.5.2 hence or otherwise determine the general term of this pattern. (4) 1.6 Consider the sequence: 3; p; 10; q; 21. The sequence has a constant second difference of 1. Determine the values of p and q. (4)

The Correct Answer and Explanation is:

Let’s solve each part step by step.

1.5.1 Calculate the value of x (Given sequence: 4, 9, x, 37…)

We are told the pattern is quadratic. In a quadratic sequence, the second differences between terms are constant.

Let’s label the terms as follows:

• T1=4T_1 = 4T1​=4
• T2=9T_2 = 9T2​=9
• T3=xT_3 = xT3​=x
• T4=37T_4 = 37T4​=37

Now calculate first differences:

• T2−T1=9−4=5T_2 – T_1 = 9 – 4 = 5T2​−T1​=9−4=5
• T3−T2=x−9T_3 – T_2 = x – 9T3​−T2​=x−9
• T4−T3=37−xT_4 – T_3 = 37 – xT4​−T3​=37−x

Now calculate second differences:

• (x−9)−5=x−14(x – 9) – 5 = x – 14(x−9)−5=x−14
• (37−x)−(x−9)=46−2x(37 – x) – (x – 9) = 46 – 2x(37−x)−(x−9)=46−2x

Since the second differences are equal: x−14=46−2xx – 14 = 46 – 2xx−14=46−2x

Solve for xxx: x−14=46−2xx+2x=46+143x=60x=20x – 14 = 46 – 2x \\ x + 2x = 46 + 14 \\ 3x = 60 \\ x = 20x−14=46−2xx+2x=46+143x=60x=20

✅ Answer for 1.5.1: 20\boxed{20}20​

1.5.2 General term of the quadratic sequence

We use the general quadratic form: Tn=an2+bn+cT_n = an^2 + bn + cTn​=an2+bn+c

We already know:

• T1=4T_1 = 4T1​=4
• T2=9T_2 = 9T2​=9
• T3=20T_3 = 20T3​=20

Substitute into the equation:

1. a(1)2+b(1)+c=4a(1)^2 + b(1) + c = 4a(1)2+b(1)+c=4 → a+b+c=4a + b + c = 4a+b+c=4
2. a(2)2+b(2)+c=9a(2)^2 + b(2) + c = 9a(2)2+b(2)+c=9 → 4a+2b+c=94a + 2b + c = 94a+2b+c=9
3. a(3)2+b(3)+c=20a(3)^2 + b(3) + c = 20a(3)2+b(3)+c=20 → 9a+3b+c=209a + 3b + c = 209a+3b+c=20

Now solve the system:

Step 1: Subtract (1) from (2): (4a+2b+c)−(a+b+c)=9−43a+b=5[Equation A](4a + 2b + c) – (a + b + c) = 9 – 4 \\ 3a + b = 5 \quad \text{[Equation A]}(4a+2b+c)−(a+b+c)=9−43a+b=5[Equation A]

Step 2: Subtract (2) from (3): (9a+3b+c)−(4a+2b+c)=20−95a+b=11[Equation B](9a + 3b + c) – (4a + 2b + c) = 20 – 9 \\ 5a + b = 11 \quad \text{[Equation B]}(9a+3b+c)−(4a+2b+c)=20−95a+b=11[Equation B]

Subtract Equation A from B: (5a+b)−(3a+b)=11−52a=6⇒a=3(5a + b) – (3a + b) = 11 – 5 \\ 2a = 6 \Rightarrow a = 3(5a+b)−(3a+b)=11−52a=6⇒a=3

Substitute into Equation A: 3(3)+b=5⇒9+b=5⇒b=−43(3) + b = 5 \Rightarrow 9 + b = 5 \Rightarrow b = -43(3)+b=5⇒9+b=5⇒b=−4

Substitute aaa and bbb into Equation 1: 3−4+c=4⇒−1+c=4⇒c=53 – 4 + c = 4 \Rightarrow -1 + c = 4 \Rightarrow c = 53−4+c=4⇒−1+c=4⇒c=5

✅ General term: Tn=3n2−4n+5\boxed{T_n = 3n^2 – 4n + 5}Tn​=3n2−4n+5​

1.6 Find values of p and q in the sequence: 3, p, 10, q, 21

We are told the second differences are constant and equal to 1.

Let’s define the terms:

• T1=3T_1 = 3T1​=3
• T2=pT_2 = pT2​=p
• T3=10T_3 = 10T3​=10
• T4=qT_4 = qT4​=q
• T5=21T_5 = 21T5​=21

Let’s write the first differences:

• T2−T1=p−3T_2 – T_1 = p – 3T2​−T1​=p−3
• T3−T2=10−pT_3 – T_2 = 10 – pT3​−T2​=10−p
• T4−T3=q−10T_4 – T_3 = q – 10T4​−T3​=q−10
• T5−T4=21−qT_5 – T_4 = 21 – qT5​−T4​=21−q

Now write the second differences:

1. (10−p)−(p−3)=13−2p(10 – p) – (p – 3) = 13 – 2p(10−p)−(p−3)=13−2p
2. (q−10)−(10−p)=q−20+p=p+q−20(q – 10) – (10 – p) = q – 20 + p = p + q – 20(q−10)−(10−p)=q−20+p=p+q−20
3. (21−q)−(q−10)=31−2q(21 – q) – (q – 10) = 31 – 2q(21−q)−(q−10)=31−2q

Since all second differences are equal:

Set 1st and 2nd equal: 13−2p=p+q−20[Equation 1]13 – 2p = p + q – 20 \quad \text{[Equation 1]}13−2p=p+q−20[Equation 1]

Set 2nd and 3rd equal: p+q−20=31−2q[Equation 2]p + q – 20 = 31 – 2q \quad \text{[Equation 2]}p+q−20=31−2q[Equation 2]

Solve Equation 2: p+q−20=31−2qp+3q=51[Equation A]p + q – 20 = 31 – 2q \\ p + 3q = 51 \quad \text{[Equation A]}p+q−20=31−2qp+3q=51[Equation A]

Now solve Equation 1: 13−2p=p+q−2033=3p+q[Equation B]13 – 2p = p + q – 20 \\ 33 = 3p + q \quad \text{[Equation B]}13−2p=p+q−2033=3p+q[Equation B]

Now solve equations A and B simultaneously:

From A: p=51−3q1p = \frac{51 – 3q}{1}p=151−3q​

Substitute into B: 3(51−3q1)+q=33153−9q+q=33153−8q=33⇒8q=120⇒q=153\left( \frac{51 – 3q}{1} \right) + q = 33 \\ 153 – 9q + q = 33 \\ 153 – 8q = 33 \Rightarrow 8q = 120 \Rightarrow q = 153(151−3q​)+q=33153−9q+q=33153−8q=33⇒8q=120⇒q=15

Substitute into A: p+3(15)=51⇒p=6p + 3(15) = 51 \Rightarrow p = 6p+3(15)=51⇒p=6

✅ Answer: p=6, q=15\boxed{p = 6,\ q = 15}p=6, q=15​

Explanation

Both parts of this question explore quadratic patterns, which are sequences with a constant second difference. Understanding these patterns allows us to fill in missing values or determine a general formula.

In question 1.5.1, we are given a quadratic sequence where some values are known and one is missing. We calculate the first and second differences to find a value for the unknown term, xxx. Since second differences are constant in a quadratic sequence, we can set two expressions for the second difference equal to each other. Solving the resulting equation yields the value of x=20x = 20x=20.

In question 1.5.2, we are asked to determine a general term for the sequence. Any quadratic pattern follows the form Tn=an2+bn+cT_n = an^2 + bn + cTn​=an2+bn+c. By substituting the known terms into this expression and solving the resulting system of equations, we determine the values of aaa, bbb, and ccc. This leads us to the general term Tn=3n2−4n+5T_n = 3n^2 – 4n + 5Tn​=3n2−4n+5.

In question 1.6, we are given another quadratic sequence with missing terms, but this time the second difference is explicitly stated as 1. Using this information, we calculate the first and second differences in terms of the unknowns ppp and qqq. Equating the second differences allows us to build a system of equations. Solving these simultaneously gives the values p=6p = 6p=6 and q=15q = 15q=15.

Recognizing the structure of quadratic sequences is key to working with these types of problems. With consistent steps—first differences, second differences, and algebraic solving—you can find missing terms and general formulas efficiently.