2.1 Given: \( 0 ;-\frac{1}{2} ;-1 ; \frac{3}{2} ; . \). 2.1 .1 Show that \( T_{n}=\frac{-n+1}{2} \) \( 2.1 .2 \quad \) Determine \( T_{10} \) \( 2.2 \quad \) Consider the following terms of the Geometric Sequence: \( 4 ; x ; 9 ; \ldots \) Determine the value of \( x \) \( \quad \begin{array}{l} \text { If the following are properties of a quadratic sequence: } \\ >T_{1}=3 \\ >5 a+b=6 \\ 2.3 .1 \text { The second difference is } 4\end{array} \)

We are given three separate problems. We will solve each one step‐by‐step. ────────────────────────────── Part 2.1 The sequence given is:   0; –½; –1; 3/2; … • 2.1.1 Show that Tₙ = (–n + 1)/2 Let’s check a few terms using Tₙ = (–n + 1)/2:   For n = 1:    T₁ = (–1 + 1)/2 = 0/2 = 0   For n = 2:    T₂ = (–2 + 1)/2 = (–1)/2 = –½   For n = 3:    T₃ = (–3 + 1)/2 = (–2)/2 = –1   For n = 4:    T₄ = (–4 + 1)/2 = (–3)/2 = –1.5, which is the same as 3/2 in absolute value but with opposite sign. Notice that the term given as “3/2” in the problem statement appears to be positive. However, if we follow the pattern from the first three terms using Tₙ = (–n+1)/2, the fourth term should be –3/2. It is possible that there might be a sign misprint in listing the terms, or the intended sequence is defined by Tₙ = (–n+1)/2. In any case, our verification shows that if the formula is assumed to be Tₙ = (–n+1)/2 then T₁, T₂, and T₃ match the given terms. • 2.1.2 Determine T₁₀ Using Tₙ = (–n+1)/2:   T₁₀ = (–10 + 1)/2 = (–9)/2 = –4.5 ────────────────────────────── Part 2.2 We are given a geometric sequence with three consecutive terms:   4; x; 9; … For a geometric sequence the ratio between successive terms is constant. Let r be the common ratio. Then:   x = 4 · r    (1)   9 = x · r    (2) Substitute (1) into (2):   9 = (4r) · r = 4r²   r² = 9/4  ⇒ r = 3/2  or r = –3/2 Now, using x = 4r, we have two choices:   If r = 3/2, then x = 4 · (3/2) = 6   If r = –3/2, then x = 4 · (–3/2) = –6 Thus, x can be 6 or –6. ────────────────────────────── Part 2.3 We are told that a quadratic sequence has the following properties:   • T₁ = 3   • 5a + b = 6   • The second difference is 4 A quadratic sequence can be written in the form:   Tₙ = a·n² + b·n + c Step 1. Use the fact that T₁ = 3:   T₁ = a(1)² + b(1) + c = a + b + c = 3    (Equation A) Step 2. The second difference of a quadratic sequence is constant and equals 2a. We are given that the second difference is 4:   2a = 4  ⇒ a = 2 Step 3. We are also given that:   5a + b = 6 Substitute a = 2:   5(2) + b = 6  ⇒ 10 + b = 6  ⇒ b = 6 – 10 = –4 Step 4. Now use Equation A to find c:   a + b + c = 3   2 + (–4) + c = 3  ⇒ (–2) + c = 3  ⇒ c = 3 + 2 = 5 Thus, the quadratic sequence is:   Tₙ = 2n² – 4n + 5 ────────────────────────────── Summary of Answers 2.1.1 The nth term of the sequence is Tₙ = (–n + 1)/2 2.1.2 T₁₀ = –9/2 (or –4.5) 2.2 For the geometric sequence 4, x, 9, … the value of x is found by solving x² = 36 so that x = 6 or x = –6 2.3 For the quadratic sequence:   a = 2, b = –4, c = 5, so Tₙ = 2n² – 4n + 5 These are the required results.