Answer
\( x = 13 \) and \( y = 65 \).
Solution
Let's solve the problems step by step.
### 9.1 Calculate \( x \) in the sequence \( 4, 9, x, 37, \ldots, j \)
A quadratic sequence can be defined by a second difference that is constant. Let's denote the terms of the sequence as follows:
- \( a_1 = 4 \)
- \( a_2 = 9 \)
- \( a_3 = x \)
- \( a_4 = 37 \)
First, we calculate the first differences:
- \( a_2 - a_1 = 9 - 4 = 5 \)
- \( a_3 - a_2 = x - 9 \)
- \( a_4 - a_3 = 37 - x \)
Now, we calculate the second differences:
- First difference between the first two first differences:
\[
(x - 9) - 5 = x - 14
\]
- Second difference between the last two first differences:
\[
(37 - x) - (x - 9) = 37 - x - x + 9 = 46 - 2x
\]
Since the second differences must be equal, we set them equal to each other:
\[
x - 14 = 46 - 2x
\]
Now, let's solve for \( x \):
\[
x + 2x = 46 + 14
\]
\[
3x = 60
\]
\[
x = 20
\]
### 9.2 Hence, or otherwise, determine the nth term of the sequence
Now that we have \( x = 20 \), the sequence is \( 4, 9, 20, 37, \ldots \).
To find the nth term, we can use the general form of a quadratic sequence:
\[
a_n = An^2 + Bn + C
\]
We can set up a system of equations using the known terms:
1. For \( n = 1 \): \( A(1^2) + B(1) + C = 4 \) → \( A + B + C = 4 \)
2. For \( n = 2 \): \( A(2^2) + B(2) + C = 9 \) → \( 4A + 2B + C = 9 \)
3. For \( n = 3 \): \( A(3^2) + B(3) + C = 20 \) → \( 9A + 3B + C = 20 \)
Now we can solve this system of equations.
Let's denote the equations as follows:
1. \( A + B + C = 4 \) (Equation 1)
2. \( 4A + 2B + C = 9 \) (Equation 2)
3. \( 9A + 3B + C = 20 \) (Equation 3)
We can eliminate \( C \) by subtracting Equation 1 from Equations 2 and 3.
Subtracting Equation 1 from Equation 2:
\[
(4A + 2B + C) - (A + B + C) = 9 - 4
\]
\[
3A + B = 5 \quad (Equation 4)
\]
Subtracting Equation 1 from Equation 3:
\[
(9A + 3B + C) - (A + B + C) = 20 - 4
\]
\[
8A + 2B = 16 \quad (Equation 5)
\]
Now we can simplify Equation 5:
\[
4A + B = 8 \quad (Equation 6)
\]
Now we have a new system of equations:
1. \( 3A + B = 5 \) (Equation 4)
2. \( 4A + B = 8 \) (Equation 6)
Subtract Equation 4 from Equation 6:
\[
(4A + B) - (3A + B) = 8 - 5
\]
\[
A = 3
\]
Substituting \( A = 3 \) back into Equation 4:
\[
3(3) + B = 5
\]
\[
9 + B = 5
\]
\[
B = -4
\]
Now substituting \( A \) and \( B \) back into Equation 1 to find \( C \):
\[
3 - 4 + C = 4
\]
\[
C = 5
\]
Thus, the nth term of the sequence is:
\[
a_n = 3n^2 - 4n + 5
\]
### 9.2 Calculate the values of \( x \) and \( y \) in the sequence \( 6, x, 26, 45, y \)
Using the same method as before, we denote the terms:
- \( b_1 = 6 \)
- \( b_2 = x \)
- \( b_3 = 26 \)
- \( b_4 = 45 \)
- \( b_5 = y \)
Calculating the first differences:
- \( b_2 - b_1 = x - 6 \)
- \( b_3 - b_2 = 26 - x \)
- \( b_4 - b_3 = 45 - 26 = 19 \)
Now, we calculate the second differences:
- First difference between the first two first differences:
\[
(26 - x) - (x - 6) = 26 - x - x + 6 = 32 - 2x
\]
- Second difference between the last two first differences:
\[
19 - (26 - x) = 19 - 26 + x = x - 7
\]
Setting the second differences equal:
\[
32 - 2x = x - 7
\]
Now, let's solve for \( x \):
\[
32 + 7 = 3x
\]
\[
39 = 3x
\]
\[
x = 13
\]
Now we have \( x = 13 \). The sequence is \( 6, 13, 26, 45, y \).
Next, we can find \( y \) by calculating the first differences again:
- \( b_1 = 6 \)
- \( b_2 = 13 \)
- \( b_3 = 26 \)
- \( b_4 = 45 \)
- \( b_5 = y \)
Calculating the first differences:
- \( b_2 - b_1 = 13 - 6 = 7 \)
- \( b_3 - b_2 = 26 - 13 = 13 \)
- \( b_4 - b_3 = 45 - 26 = 19 \)
- \( b_5 - b_4 = y - 45 \
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