The Sieve of Sundaram is an infinite table of arithmetic sequences. The terms in the first 4 rows and the first 4 columns of the table are shown below. | | | | | |---|---|---|---| | 4 | 7 | 10 | 13 | | 7 | 12 | 17 | 22 | | 10 | 17 | 24 | 31 | | 13 | 22 | 31 | 40 | (i) Find the difference between the sums of the first 45 terms in the first two rows. (ii) Find the number which is in the 60th row and 70th column of the table. (b) The first two terms of a sequence are a₁ = 4 and a₂ = 2. The general term is defined by aₙ = aₙ₋₁ − aₙ₋₂ when n ≥ 3. Write out the next 6 terms of the sequence and hence find the value of a₂₀₁₉.

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The Sieve of Sundaram is an infinite table of arithmetic sequences - Leaving Cert Mathematics - Question 5 - 2018

Question 5

The Sieve of Sundaram is an infinite table of arithmetic sequences. The terms in the first 4 rows and the first 4 columns of the table are shown below. | | | ... show full transcript

Worked Solution & Example Answer:The Sieve of Sundaram is an infinite table of arithmetic sequences - Leaving Cert Mathematics - Question 5 - 2018

Step 1

Find the difference between the sums of the first 45 terms in the first two rows.

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Answer

To find the sums of the first 45 terms in each row, we start by determining the number of terms in these rows.

Row 1 contains 4 terms: 4, 7, 10, 13.
Row 2 also contains 4 terms: 7, 12, 17, 22.

Using the formula for the sum of an arithmetic sequence, we can calculate:

Sum of Row 1 (S₁):
- First term (a₁) = 4
- Common difference (d) = 3
- Number of terms (n) = 4
- Using the formula:
$S_n = rac{n}{2} (2a + (n-1)d)$

We have:

$S₁ = rac{4}{2} (2 imes 4 + (4-1) imes 3) = 2(8 + 9) = 2 imes 17 = 34$
Sum of Row 2 (S₂):
- First term (a₂) = 7
- Common difference (d) = 5
- Number of terms (n) = 4
- Using the same formula:
$S₂ = rac{4}{2} (2 imes 7 + (4-1) imes 5) = 2(14 + 15) = 2 imes 29 = 58$
Difference:
- The difference between the sums is:
$|S₁ - S₂| = |34 - 58| = 24$

Step 2

Find the number which is in the 60th row and 70th column of the table.

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Answer

To find the number in the 60th row and 70th column, we first need to determine the structure of the table.

The formula for determining the number in any row (r) and column (c) is:

$T(r, c) = 4 + (r - 1) + (c - 1)(r + c - 2)$
Substituting r = 60 and c = 70:

$T(60, 70) = 4 + (60 - 1) + (70 - 1)(60 + 70 - 2)$

Simplifying yields:

$= 4 + 59 + 69(128)$

Performing calculations gives:

$= 4 + 59 + 8832$

Therefore, the value is:

$T(60, 70) = 8895$

Step 3

Write out the next 6 terms of the sequence and hence find the value of a₂₀₁₉.

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Answer

Given the first two terms a₁ = 4 and a₂ = 2, we can calculate the next terms using the provided general term relation:

Next Terms Calculation:
- a₃ = a₂ - a₁ = 2 - 4 = -2
- a₄ = a₃ - a₂ = -2 - 2 = -4
- a₅ = a₄ - a₃ = -4 - (-2) = -2
- a₆ = a₅ - a₄ = -2 - (-4) = 2
- a₇ = a₆ - a₅ = 2 - (-2) = 4
- a₈ = a₇ - a₆ = 4 - 2 = 2

Now we observe a repeating pattern: 4, 2, -2, -4.

Finding a₂₀₁₉:
- The sequence consists of 6 terms: 4, 2, -2, -4, -2, 2.
- Notice the pattern every 6 terms: we can calculate a₂₀₁₉ considering the periodicity of 6.
- 2019 mod 6 = 3 yields a value of a₃ = -2.
- Therefore, a₂₀₁₉ = -2.

The Sieve of Sundaram is an infinite table of arithmetic sequences - Leaving Cert Mathematics - Question 5 - 2018

Worked Solution & Example Answer:The Sieve of Sundaram is an infinite table of arithmetic sequences - Leaving Cert Mathematics - Question 5 - 2018

Find the difference between the sums of the first 45 terms in the first two rows.

Find the number which is in the 60th row and 70th column of the table.

Write out the next 6 terms of the sequence and hence find the value of a₂₀₁₉.

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