Investigate the relationship between the common differeace, \( d \), end the Difference, \( D \), between (the product of the first and tho third tems) and (the square of the middle term) of any three consecutive numbers of a linear sequence. NOTE: - If tha sequenco is given by: \( a ; b ; c ; \)..the Difference \( (D)=b^{2}-a \times c \) Step 1 - Use IIVE different linear sequences with common difference, \( d \) of \( I \) and calculate the difference, \( D \) as stated above. (4) Sten 2 - Write dówn your conjecture in words (2) Step 3 - Repeat step 1: change the common difficience, \( d \) between the ternis to 2. (4) Step 4 - Repeat step I chasge the common difference, d between the terms to 3. (4) Step 5 - Repeat step 1 change the common difference, d between the terms to a number of your choice (NOT 1,2 or 3 ). (4) Step 6 - Write down your conjecture in words. (2) Step 7 - Is this trua in general? Prove your conjecture. (5) Step 8 - Overall presentation and varicty of sequences used

Below is a detailed, step‐by‐step investigation of the relationship between the common difference, \( d \), and the difference, \( D \), where \[ D = b^2 - a \times c \] for three consecutive terms \( a \), \( b \), and \( c \) in an arithmetic (linear) sequence. --- ## Step 1: Five Linear Sequences with Common Difference \( d = 1 \) An arithmetic sequence with \( d = 1 \) can be written as: \[ a,\quad a+1,\quad a+2 \] The difference is given by \[ D = (a+1)^2 - a(a+2) \] Let’s expand and simplify: \[ \begin{aligned} D &= (a^2 + 2a + 1) - (a^2 + 2a) \\ &= 1. \end{aligned} \] Thus, regardless of the starting value \( a \), \( D=1 \). We now list five different examples: 1. For \( a = 1 \): Sequence: \( 1,\, 2,\, 3 \) \(\Rightarrow D = 1\). 2. For \( a = 2 \): Sequence: \( 2,\, 3,\, 4 \) \(\Rightarrow D = 1\). 3. For \( a = 0 \): Sequence: \( 0,\, 1,\, 2 \) \(\Rightarrow D = 1\). 4. For \( a = -3 \): Sequence: \( -3,\, -2,\, -1 \) \(\Rightarrow D = 1\). 5. For \( a = 10 \): Sequence: \( 10,\, 11,\, 12 \) \(\Rightarrow D = 1\). --- ## Step 2: Conjecture for \( d = 1 \) **Conjecture:** For any three consecutive numbers in an arithmetic sequence with common difference \( d = 1 \), the difference between the square of the middle term and the product of the first and third terms is always \( 1 \). --- ## Step 3: Five Linear Sequences with Common Difference \( d = 2 \) For \( d = 2 \), any arithmetic sequence is: \[ a,\quad a+2,\quad a+4. \] The difference is: \[ D = (a+2)^2 - a(a+4) \] Computing this: \[ \begin{aligned} D &= (a^2 + 4a + 4) - (a^2 + 4a) \\ &= 4. \end{aligned} \] Thus, \( D=4 \) regardless of \( a \). Examples: 1. For \( a = 1 \): Sequence: \( 1,\, 3,\, 5 \) \(\Rightarrow D = 4\). 2. For \( a = 0 \): Sequence: \( 0,\, 2,\, 4 \) \(\Rightarrow D = 4\). 3. For \( a = -2 \): Sequence: \( -2,\, 0,\, 2 \) \(\Rightarrow D = 4\). 4. For \( a = 5 \): Sequence: \( 5,\, 7,\, 9 \) \(\Rightarrow D = 4\). 5. For \( a = 10 \): Sequence: \( 10,\, 12,\, 14 \) \(\Rightarrow D = 4\). --- ## Step 4: Five Linear Sequences with Common Difference \( d = 3 \) For \( d = 3 \), the sequence becomes: \[ a,\quad a+3,\quad a+6. \] Calculate the difference: \[ D = (a+3)^2 - a(a+6) \] Simplify: \[ \begin{aligned} D &= (a^2 + 6a + 9) - (a^2 + 6a) \\ &= 9. \end{aligned} \] Thus, \( D=9 \) regardless of \( a \). Example sequences: 1. For \( a = 1 \): Sequence: \( 1,\, 4,\, 7 \) \(\Rightarrow D = 9\). 2. For \( a = 0 \): Sequence: \( 0,\, 3,\, 6 \) \(\Rightarrow D = 9\). 3. For \( a = -1 \): Sequence: \( -1,\, 2,\, 5 \) \(\Rightarrow D = 9\). 4. For \( a = 3 \): Sequence: \( 3,\, 6,\, 9 \) \(\Rightarrow D = 9\). 5. For \( a = 10 \): Sequence: \( 10,\, 13,\, 16 \) \(\Rightarrow D = 9\). --- ## Step 5: Five Linear Sequences with a Different Common Difference \( d \) (Chosen \( d = 5 \)) For \( d = 5 \), the sequence is: \[ a,\quad a+5,\quad a+10. \] The difference is: \[ D = (a+5)^2 - a(a+10) \] Simplify it: \[ \begin{aligned} D &= (a^2 + 10a + 25) - (a^2 + 10a) \\ &= 25. \end{aligned} \] Thus, \( D=25 \) regardless of \( a \). Example sequences: 1. For \( a = 1 \): Sequence: \( 1,\, 6,\, 11 \) \(\Rightarrow D = 25\). 2. For \( a = 0 \): Sequence: \( 0,\, 5,\, 10 \) \(\Rightarrow D = 25\). 3. For \( a = -4 \): Sequence: \( -4,\, 1,\, 6 \) \(\Rightarrow D = 25\). 4. For \( a = 7 \): Sequence: \( 7,\, 12,\, 17 \) \(\Rightarrow D = 25\). 5. For \( a = 10 \): Sequence: \( 10,\, 15,\, 20 \) \(\Rightarrow D = 25\). --- ## Step 6: Conjecture in Words **Conjecture:** For any three consecutive terms of an arithmetic sequence with common difference \( d \), the difference between the square of the middle term and the product of the first and third terms is always equal to the square of the common difference. In other words, \[ D = d^2. \] --- ## Step 7: Proof of the Conjecture Consider any arithmetic sequence where the three consecutive terms are: \[ a,\quad a+d,\quad a+2d. \] Then the difference is: \[ D = (a+d)^2 - a(a+2d). \] Expand and simplify: \[ \begin{aligned} (a+d)^2 &= a^2 + 2ad + d^2, \\ a(a+2d) &= a^2 + 2ad. \end{aligned} \] Substitute back into the expression for \( D \): \[ \begin{aligned} D &= (a^2 + 2ad + d^2) - (a^2 + 2ad) \\ &= d^2. \end{aligned} \] This proves that for any three consecutive terms in an arithmetic sequence, the difference \( D \) is always equal to \( d^2 \). --- ## Step 8: Overall Presentation and Variety of Sequences Used In this investigation we: - **Selected five different sequences** for each chosen value of \( d \) (namely \( d = 1, 2, 3 \), and a chosen \( d = 5 \)) to calculate \( D \). - **Found consistent results** across all examples: - With \( d = 1 \), \( D = 1 \). - With \( d = 2 \), \( D = 4 \). - With \( d = 3 \), \( D = 9 \). - With \( d = 5 \), \( D = 25 \). - **Derived and proved the general result** that for three consecutive terms \( a \), \( a+d \), \( a+2d \), the difference is always: \[ D = d^2. \] Thus, the relationship between the common difference \( d \) and the difference \( D \) is that \( D \) is exactly the square of the common difference, i.e., \( D = d^2 \). --- This completes the investigation step by step.