1. Consider the pattern: 10; 7; 4; 1; ... 1.1 Determine the formula $T_n$, the general term of the sequence. Step 1: Identify the pattern type. This is an arithmetic sequence because the difference between terms is constant. Step 2: Calculate the common difference $d$. $$d = 7 - 10 = -3$$ Step 3: Use the arithmetic sequence formula: $$T_n = a + (n-1)d$$ where $a = 10$ (first term) and $d = -3$. Step 4: Substitute values: $$T_n = 10 + (n-1)(-3)$$ $$T_n = 10 - 3(n-1)$$ $$T_n = 10 - 3n + 3$$ $$T_n = 13 - 3n$$ 1.2 Which term in the sequence is equal to -113? Step 1: Set $T_n = -113$ and solve for $n$. $$13 - 3n = -113$$ Step 2: Subtract 13 from both sides: $$-3n = -113 - 13$$ $$-3n = -126$$ Step 3: Divide both sides by $\cancel{-3}$: $$n = \frac{-126}{\cancel{-3}} = 42$$ Answer: The 42nd term is -113. 2. The sequence 1; 6; 15; 28; ... is given. 2.1 Determine a formula for the nth term of the sequence. Step 1: Check if the sequence is arithmetic. Differences: $6-1=5$, $15-6=9$, $28-15=13$ (not constant) Step 2: Check second differences: $9-5=4$, $13-9=4$ (constant second difference) Step 3: Since second difference is constant, it's a quadratic sequence. General form: $$T_n = an^2 + bn + c$$ Step 4: Use the first three terms to form equations: For $n=1$: $a(1)^2 + b(1) + c = 1$ => $a + b + c = 1$ For $n=2$: $4a + 2b + c = 6$ For $n=3$: $9a + 3b + c = 15$ Step 5: Subtract first equation from second and third: $$ (4a + 2b + c) - (a + b + c) = 6 - 1 $$ $$ 3a + b = 5 $$ $$ (9a + 3b + c) - (a + b + c) = 15 - 1 $$ $$ 8a + 2b = 14 $$ Step 6: Solve system: From $3a + b = 5$, express $b = 5 - 3a$ Substitute into $8a + 2b = 14$: $$8a + 2(5 - 3a) = 14$$ $$8a + 10 - 6a = 14$$ $$2a + 10 = 14$$ $$2a = 4$$ $$a = 2$$ Step 7: Find $b$: $$b = 5 - 3(2) = 5 - 6 = -1$$ Step 8: Find $c$ using $a + b + c = 1$: $$2 - 1 + c = 1$$ $$c = 0$$ Step 9: Final formula: $$T_n = 2n^2 - n$$ 2.2 Determine the value of term number 100. Step 1: Substitute $n=100$: $$T_{100} = 2(100)^2 - 100 = 2(10000) - 100 = 20000 - 100 = 19900$$ 3. Calculate the value of $x$ in the quadratic sequence: 1; 5; x; 19; ... Step 1: Identify the sequence terms: $$T_1 = 1, T_2 = 5, T_3 = x, T_4 = 19$$ Step 2: Calculate first differences: $$5 - 1 = 4$$ $$x - 5 = ?$$ $$19 - x = ?$$ Step 3: Calculate second differences: Second difference between terms 1 and 2: $$(x - 5) - 4 = (x - 9)$$ Second difference between terms 2 and 3: $$(19 - x) - (x - 5) = 19 - x - x + 5 = 24 - 2x$$ Step 4: For quadratic sequences, second differences are constant: $$x - 9 = 24 - 2x$$ Step 5: Solve for $x$: $$x - 9 = 24 - 2x$$ $$x + 2x = 24 + 9$$ $$3x = 33$$ $$x = 11$$ 4. Solve the equation: $$-4x - 5 = 19 - x$$ Step 1: Add $4x$ to both sides: $$-5 = 19 - x + 4x$$ $$-5 = 19 + 3x$$ Step 2: Subtract 19 from both sides: $$-5 - 19 = 3x$$ $$-24 = 3x$$ Step 3: Divide both sides by $\cancel{3}$: $$x = \frac{-24}{\cancel{3}} = -8$$