Investigate the reiationship between the common differcace, \( d \), and the Difference, \( D \), between (the produet of the first and the third termis) and (the square of the middle term) of any three consecutive cumbers of a linear sequence. NOTE: - If the sequence is given by \( a ; b ; c \), .. the Difference \( (D)=b^{2}-a \times c \) Step 1 - Use FIVE differect linear sequences with common difference, \( d \), of \( I \) and calculate the diffirence, \( D \) a 5 stated above. (4) Step 2 - Write dōw your conjecture in words. (2) Step 3 - Repeat şep 1: change the common difference, d between the terms to 2 . (4) Step 4 - Kepeat step 1. change the common differcace, d between the terms to 3 . (4) Step 5 - Repeat step ! change the common difference, \( d \) berween the terms to a number of your choiec (NOT 1, 2 or 3). (4) Step 6 - Write down your conjecture in words. (2) Step 7 - Is this true in general? Prove your conjecture. (5) Step 8 - Overall presentation and variety of sequences used (10) [39]
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The Deep Dive
To get started on your investigation into the relationship between the common difference, \(d\), and the difference \(D = b^2 - ac\), let's dig into some calculations first: **Step 1: Common difference \(d = 1\)** 1. Sequence: \(1, 2, 3\) → \(D = 2^2 - (1 \times 3) = 4 - 3 = 1\) 2. Sequence: \(4, 5, 6\) → \(D = 5^2 - (4 \times 6) = 25 - 24 = 1\) 3. Sequence: \(7, 8, 9\) → \(D = 8^2 - (7 \times 9) = 64 - 63 = 1\) 4. Sequence: \(10, 11, 12\) → \(D = 11^2 - (10 \times 12) = 121 - 120 = 1\) 5. Sequence: \(13, 14, 15\) → \(D = 14^2 - (13 \times 15) = 196 - 195 = 1\) **Step 2:** When the common difference \(d\) is 1 in any linear sequence, the difference \(D\) between the product of the first and third terms and the square of the middle term always equals 1. **Step 3: Common difference \(d = 2\)** 1. Sequence: \(2, 4, 6\) → \(D = 4^2 - (2 \times 6) = 16 - 12 = 4\) 2. Sequence: \(10, 12, 14\) → \(D = 12^2 - (10 \times 14) = 144 - 140 = 4\) 3. Sequence: \(18, 20, 22\) → \(D = 20^2 - (18 \times 22) = 400 - 396 = 4\) 4. Sequence: \(26, 28, 30\) → \(D = 28^2 - (26 \times 30) = 784 - 780 = 4\) 5. Sequence: \(34, 36, 38\) → \(D = 36^2 - (34 \times 38) = 1296 - 1292 = 4\) **Step 4:** For a common difference of 2, \(D\) appears to always be 4. **Step 5: Common difference \(d = 3\)** 1. Sequence: \(3, 6, 9\) → \(D = 6^2 - (3 \times 9) = 36 - 27 = 9\) 2. Sequence: \(12, 15, 18\) → \(D = 15^2 - (12 \times 18) = 225 - 216 = 9\) 3. Sequence: \(21, 24, 27\) → \(D = 24^2 - (21 \times 27) = 576 - 567 = 9\) 4. Sequence: \(30, 33, 36\) → \(D = 33^2 - (30 \times 36) = 1089 - 1080 = 9\) 5. Sequence: \(39, 42, 45\) → \(D = 42^2 - (39 \times 45) = 1764 - 1755 = 9\) **Step 6:** The pattern continues; with a common difference of \(3\), \(D\) consistently results in \(9\). **Step 7:** In general, for a common difference \(d\), if you set \(n\) as the number of the first term in the sequence: - The terms can be represented as \(n\), \(n+d\), \(n+2d\). - Calculating \(D\) gives us \(D = (n+d)^2 - n(n+2d)\). - So we can simplify: \(D = n^2 + 2nd + d^2 - n^2 - 2nd = d^2\). - Therefore, the difference \(D\) is equal to the square of the common difference \(d\) regardless of the starting point in the linear sequence. **Step 8:** Make sure your overall presentation is clear, with neatly labeled steps and well-organized calculations. Use a variety of sequences to illustrate your points, and integrate fun visuals or color coding to engage
