Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Our community is ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
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].
### 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].
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.
