Question
upstudy study bank question image url

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]

Ask by Pearson Bush. in South Africa
Mar 07,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

For any three consecutive terms in a linear sequence with a common difference \( d \), the difference \( D \) between the square of the middle term and the product of the first and third terms is always \( d^2 \).

Solution

To investigate the relationship between the common difference \( d \) and the difference \( D \) defined as \( D = b^2 - a \times c \) for three consecutive terms of a linear sequence, we will follow the steps outlined in the problem. ### Step 1: Common Difference \( d = 1 \) Let's consider five different linear sequences with a common difference of \( d = 1 \): 1. Sequence: \( 1, 2, 3 \) - \( a = 1, b = 2, c = 3 \) - \( D = 2^2 - 1 \times 3 = 4 - 3 = 1 \) 2. Sequence: \( 4, 5, 6 \) - \( a = 4, b = 5, c = 6 \) - \( D = 5^2 - 4 \times 6 = 25 - 24 = 1 \) 3. Sequence: \( 7, 8, 9 \) - \( a = 7, b = 8, c = 9 \) - \( D = 8^2 - 7 \times 9 = 64 - 63 = 1 \) 4. Sequence: \( 10, 11, 12 \) - \( a = 10, b = 11, c = 12 \) - \( D = 11^2 - 10 \times 12 = 121 - 120 = 1 \) 5. Sequence: \( 13, 14, 15 \) - \( a = 13, b = 14, c = 15 \) - \( D = 14^2 - 13 \times 15 = 196 - 195 = 1 \) ### Step 2: Conjecture for \( d = 1 \) For a linear sequence with a common difference of \( 1 \), the difference \( D \) between the product of the first and third terms and the square of the middle term is always \( 1 \). ### Step 3: Common Difference \( d = 2 \) Now, let's consider five different linear sequences with a common difference of \( d = 2 \): 1. Sequence: \( 1, 3, 5 \) - \( a = 1, b = 3, c = 5 \) - \( D = 3^2 - 1 \times 5 = 9 - 5 = 4 \) 2. Sequence: \( 2, 4, 6 \) - \( a = 2, b = 4, c = 6 \) - \( D = 4^2 - 2 \times 6 = 16 - 12 = 4 \) 3. Sequence: \( 3, 5, 7 \) - \( a = 3, b = 5, c = 7 \) - \( D = 5^2 - 3 \times 7 = 25 - 21 = 4 \) 4. Sequence: \( 4, 6, 8 \) - \( a = 4, b = 6, c = 8 \) - \( D = 6^2 - 4 \times 8 = 36 - 32 = 4 \) 5. Sequence: \( 5, 7, 9 \) - \( a = 5, b = 7, c = 9 \) - \( D = 7^2 - 5 \times 9 = 49 - 45 = 4 \) ### Step 4: Conjecture for \( d = 2 \) For a linear sequence with a common difference of \( 2 \), the difference \( D \) is always \( 4 \). ### Step 5: Common Difference \( d = 3 \) Now, let's consider five different linear sequences with a common difference of \( d = 3 \): 1. Sequence: \( 1, 4, 7 \) - \( a = 1, b = 4, c = 7 \) - \( D = 4^2 - 1 \times 7 = 16 - 7 = 9 \) 2. Sequence: \( 2, 5, 8 \) - \( a = 2, b = 5, c = 8 \) - \( D = 5^2 - 2 \times 8 = 25 - 16 = 9 \) 3. Sequence: \( 3, 6, 9 \) - \( a = 3, b = 6, c = 9 \) - \( D = 6^2 - 3 \times 9 = 36 - 27 = 9 \) 4. Sequence: \( 4, 7, 10 \) - \( a = 4, b = 7, c = 10 \) - \( D = 7^2 - 4 \times 10 = 49 - 40 = 9 \) 5. Sequence: \( 5, 8, 11 \) - \( a = 5, b = 8, c = 11 \) - \( D = 8^2 - 5 \times 11 = 64 - 55 = 9 \) ### Step 6: Conjecture for \( d = 3 \) For a linear sequence with a common difference of \( 3 \), the difference \( D \) is always \( 9 \). ### Step 7: Common Difference \( d = 4 \) Now, let's consider five different linear sequences with a common difference of \( d = 4 \): 1. Sequence: \( 1, 5, 9 \) - \( a = 1, b = 5, c = 9 \) - \( D = 5^2 - 1 \times 9 = 25 - 9 = 16 \) 2. Sequence: \( 2, 6, 10 \) - \( a = 2, b = 6, c = 10 \) - \( D = 6^2 - 2 \times 10 = 36 - 20 = 16 \) 3. Sequence: \( 3, 7, 11 \) - \( a = 3, b = 7, c = 11 \) - \( D = 7^2 - 3 \times 11 = 49 - 33 = 16 \) 4. Sequence: \( 4, 8, 12 \) - \( a = 4, b = 8, c = 12 \) - \( D = 8^2 - 4 \times 12 = 64 - 48 = 16 \) 5. Sequence: \( 5,

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

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

Related Questions

Latest Algebra Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy