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QUESTION 2

The first four terms of a quadratic sequence are [tex]6, 5 + x, -6, \text{ and } 6x[/tex], and the general term is given by [tex]T_n = a n^2 + b n + c[/tex].

2.1 Show that the value of [tex]x = -3[/tex].

2.2 Determine the values of [tex]a, b[/tex], and [tex]c[/tex].



Answer :

GinnyAnswer
To find the terms of the quadratic sequence given by [tex]\( T_n = an^2 + bn + c \)[/tex], we are given that the first four terms are [tex]\( T_1 = 6 \)[/tex], [tex]\( T_2 = 5 + x \)[/tex], [tex]\( T_3 = -6 \)[/tex], and [tex]\( T_4 = 6x \)[/tex].

### 2.1 Show that the value of [tex]\( x = -3 \)[/tex]

To find [tex]\( x \)[/tex], we need to ensure that the sequence follows a quadratic pattern [tex]\( T_n = an^2 + bn + c \)[/tex].

Write the sequence's terms as:

[tex]\[ T_1 = a(1)^2 + b(1) + c = a + b + c \][/tex]

[tex]\[ T_2 = a(2)^2 + b(2) + c = 4a + 2b + c \][/tex]

[tex]\[ T_3 = a(3)^2 + b(3) + c = 9a + 3b + c \][/tex]

[tex]\[ T_4 = a(4)^2 + b(4) + c = 16a + 4b + c \][/tex]

We know the specific values of these terms:

[tex]\[ T_1 = 6 \][/tex]

[tex]\[ T_2 = 5 + x \][/tex]

[tex]\[ T_3 = -6 \][/tex]

[tex]\[ T_4 = 6x \][/tex]

We have the following system of equations:

1. [tex]\( a + b + c = 6 \)[/tex]
2. [tex]\( 4a + 2b + c = 5 + x \)[/tex]
3. [tex]\( 9a + 3b + c = -6 \)[/tex]
4. [tex]\( 16a + 4b + c = 6x \)[/tex]

### Step 1: Determine [tex]\( x \)[/tex]

Let's use consecutive differences, which define the consistency of quadratic sequences:

[tex]\[ \Delta T_n = T_{n+1} - T_n \][/tex]
[tex]\[ \Delta T_1 = T_2 - T_1 = (5 + x) - 6 = x - 1 \][/tex]
[tex]\[ \Delta T_2 = T_3 - T_2 = -6 - (5 + x) = -11 - x \][/tex]
[tex]\[ \Delta T_3 = T_4 - T_3 = 6x - (-6) = 6x + 6 \][/tex]

Now, the second differences should be consistent:

Second difference:
[tex]\[ \Delta^2 T_n = \Delta T_{n+1} - \Delta T_n \][/tex]
[tex]\[ \Delta^2 T_1 = \Delta T_2 - \Delta T_1 = (-11 - x) - (x - 1) = -11 - x - x + 1 = -10 - 2x \][/tex]
[tex]\[ \Delta^2 T_2 = \Delta T_3 - \Delta T_2 = (6x + 6) - (-11 - x) = 6x + 6 + 11 + x = 7x + 17 \][/tex]

For the sequence to be quadratic, the second differences must be constant:

[tex]\[ -10 - 2x = 7x + 17 \][/tex]

Solve for [tex]\( x \)[/tex]:

[tex]\[ -10 - 2x = 7x + 17 \][/tex]
[tex]\[ -10 - 17 = 7x + 2x \][/tex]
[tex]\[ -27 = 9x \][/tex]
[tex]\[ x = -3 \][/tex]

Thus, [tex]\( x \)[/tex] must be [tex]\( -3 \)[/tex].

### 2.2 Determine the values of [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex]

Now replace [tex]\( x \)[/tex] with [tex]\( -3 \)[/tex]:

[tex]\[ T_2 = 5 + (-3) = 2 \][/tex]
[tex]\[ T_4 = 6(-3) = -18 \][/tex]

We have updated equations:
1. [tex]\( a + b + c = 6 \)[/tex]
2. [tex]\( 4a + 2b + c = 2 \)[/tex]
3. [tex]\( 9a + 3b + c = -6 \)[/tex]
4. [tex]\( 16a + 4b + c = -18 \)[/tex]

### Solve the system of equations

Subtract equation 1 from equation 2:

[tex]\[ (4a + 2b + c) - (a + b + c) = 2 - 6 \][/tex]
[tex]\[ 3a + b = -4 \quad \text{(Equation 5)} \][/tex]

Subtract equation 2 from equation 3:

[tex]\[ (9a + 3b + c) - (4a + 2b + c) = -6 - 2 \][/tex]
[tex]\[ 5a + b = -8 \quad \text{(Equation 6)} \][/tex]

Subtract equation 3 from equation 4:

[tex]\[ (16a + 4b + c) - (9a + 3b + c) = -18 - (-6) \][/tex]
[tex]\[ 7a + b = -12 \quad \text{(Equation 7)} \][/tex]

Now, solve equations 5 and 6:

Subtract equation 5 from equation 6:

[tex]\[ (5a + b) - (3a + b) = -8 - (-4) \][/tex]
[tex]\[ 2a = -4 \][/tex]
[tex]\[ a = -2 \][/tex]

Substitute [tex]\( a = -2 \)[/tex] into equation 5:

[tex]\[ 3(-2) + b = -4 \][/tex]
[tex]\[ -6 + b = -4 \][/tex]
[tex]\[ b = 2 \][/tex]

Substitute [tex]\( a = -2 \)[/tex] and [tex]\( b = 2 \)[/tex] into equation 1:

[tex]\[ -2 + 2 + c = 6 \][/tex]
[tex]\[ c = 6 \][/tex]

Thus, the values are [tex]\( a = -2 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = 6 \)[/tex].