Given the following quadratic number pattern: 5 ; -4 ; -19 ; -40 ; … 1.1 Determine the constant second difference of the sequence - NSC Mathematics - Question 2 - 2017 - Paper 1

Given the second difference of -6, we can use the formula for quadratic sequences:

$T_n = an^2 + bn + c$

Where:

• The second difference indicates that 2a = -6, hence, a = -3.
• Using the term values, we can conclude:
  1. T_1 = 5: (5 = -3(1)^2 + b(1) + c)
  2. T_2 = -4: (-4 = -3(2)^2 + b(2) + c)
  3. T_3 = -19: (-19 = -3(3)^2 + b(3) + c)

Those equations yield:

• Solving simultaneously gives:
• (b = 0, c = 8)

Thus, the nth term is:

$T_n = -3n^2 + 8$

To find the term where ( T_n = -25 939 ):

Set up the equation:

$-3n^2 + 8 = -25 939$

Rearranging gives:

$-3n^2 = -25 939 - 8 \Rightarrow -3n^2 = -25 947 \Rightarrow n^2 = \frac{25 947}{3} \approx 8 649$

Taking the square root gives:

( n \approx 93 )

Thus, the 93rd term of the pattern is -25 939.

Using the general term for the arithmetic sequence:

$T_n = 2k - 7, k + 8, 2k - 1$

Find the common difference:

• Find k: From ( k + 8 - (2k - 7) = 15 \Rightarrow k = 12 )
• Substitute k into the sequence terms gives:
  • First term = 2(12) - 7 = 17.
  • Common difference, ( d = 3 )

Thus, the 15th term:

$T_{15} = 17 + (15 - 1) * 3 = 17 + 42 = 59$

To calculate the sum of the first 30 terms, use the sum formula:

$S_n = \frac{n}{2}(T_1 + T_n)$

Where:

• n = 30, T_1 = 17, and T_n = T_{30} = 17 + (30 - 1)*3 = 17 + 87 = 104

Calculating:

$S_{30} = \frac{30}{2}(17 + 104) = 15 * 121 = 1,815$