Given the following quadratic sequence: -2; 0; 3; 7; .. - NSC Mathematics - Question 2 - 2016 - Paper 1

To find the n^th term, T_n, we can use the formula for a quadratic sequence, which can be represented generally as:

$T_n = an^2 + bn + c$

1. From the sequence:

   • T₁ = -2
   • T₂ = 0
   • T₃ = 3
   • T₄ = 7

2. Using these values to create a system of equations, we get:

   • $a(1^2) + b(1) + c = -2$ (1)
   • $a(2^2) + b(2) + c = 0$ (2)
   • $a(3^2) + b(3) + c = 3$ (3)

3. Solving these equations, we find:

   • $a = 1$
   • $b = -2$
   • $c = -3$

Thus, the n^th term can be expressed as:

$T_n = n^2 - 2n - 3$

To find which term equals 322, we set T_n equal to 322:

$n^2 - 2n - 3 = 322$

This simplifies to:

$n^2 - 2n - 325 = 0$

Using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

• $a = 1$
• $b = -2$
• $c = -325$

This gives:

$n = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-325)}}{2(1)} = \frac{2 \pm \sqrt{4 + 1300}}{2} = \frac{2 \pm \sqrt{1304}}{2}$

Approximately, this leads to:

• $n \approx 25$ and $n \approx -26$
• Thus, the term is T₂₅ = 322.

Given:

• An arithmetic sequence where the second term T₂ = 8 and the fifth term T₅ = 10.

Using the formula for the n^th term of an arithmetic sequence, we have:

$T_n = a + (n-1)d$

Where:

• $a$ is the first term
• $d$ is the common difference.

For T₂:

• $T_2 = a + d = 8$ (1)

For T₅:

• $T_5 = a + 4d = 10$ (2)

By solving equations (1) and (2):

1. From (1): $a = 8 - d$
2. Substitute a into (2): $8 - d + 4d = 10$

ightarrow d = \frac{2}{3}$$ Therefore, the common difference is **$rac{2}{3}$**.

The sum of an arithmetic series can be computed using the formula:

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

Where:

• $T_1$ is the first term, and $T_n$ is the n^th term.

First, we need to determine T₁:

• From our earlier results: $T_1 = T_2 - d = 8 - \frac{2}{3} = \frac{22}{3}$

Next, we calculate T₅₀:

• Using $d$: $T_{50} = T_1 + (50 - 1)d = \frac{22}{3} + 49 \cdot \frac{2}{3} = \frac{22}{3} + \frac{98}{3} = \frac{120}{3} = 40$

Now plug these into the sum formula:

$S_{50} = \frac{50}{2} \left( \frac{22}{3} + 40 \right) = 25 \left( \frac{22}{3} + \frac{120}{3} \right) = 25 \cdot \frac{142}{3} = \frac{3550}{3}$

Thus, the sum of the first 50 terms is rac{3550}{3}.

Using the same sum formula for an arithmetic sequence:

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

We already established:

• $T_1 = \frac{22}{3}$
• The number of terms $n = 2$ (for the first two terms).
• We also need to find T₂ which is 8.

Hence, we calculate:

$S_2 = \frac{2}{2}(T_1 + T_2) = 1 \cdot \left( \frac{22}{3} + 8 \right) = \frac{22}{3} + \frac{24}{3} = \frac{46}{3}$

The sum of the first terms is therefore rac{46}{3}.