QUESTION 3 The first three terms of au arithmetic sequence are \( 2 p-3 ; p+5 ; 2 p+7 \). 3.1 Deternine the ralue(s) of \( p \). (3) 3.2 Calculate the sum of the first 120 terms. (3) 3.3 The following pattern is true for the arithmetic sequence above: \[ \begin{array}{l} T_{1}+T_{4}=T_{2}+T_{3} \\ T_{5}+T_{8}=T_{6}+T_{7} \\ T_{9}+T_{12}=T_{10}+T_{11} \\ \therefore T_{k}+T_{k+3}=T_{x}+T_{y} \end{array} \] 3.3.1 Write down the values of \( x \) and \( y \) in terms of \( k \). (2) 3.32 Hence. calculate the ralue of \( T_{x}+T_{y} \) in terms of \( k \) in simplest form (4)

Let's tackle each part of **Question 3** step by step. --- ### **3.1 Determine the value(s) of \( p \).** Given the first three terms of an arithmetic sequence: - \( T_1 = 2p - 3 \) - \( T_2 = p + 5 \) - \( T_3 = 2p + 7 \) In an arithmetic sequence, the difference between consecutive terms is constant. Therefore: \[ T_2 - T_1 = T_3 - T_2 \] Substituting the given terms: \[ (p + 5) - (2p - 3) = (2p + 7) - (p + 5) \] Simplify both sides: \[ -p + 8 = p + 2 \] Combine like terms: \[ -2p = -6 \quad \Rightarrow \quad p = 3 \] **Answer:** \( p = 3 \) --- ### **3.2 Calculate the sum of the first 120 terms.** First, determine the common difference \( d \) and the first term \( a_1 \) using \( p = 3 \): \[ a_1 = 2p - 3 = 2(3) - 3 = 3 \] \[ a_2 = p + 5 = 3 + 5 = 8 \quad \Rightarrow \quad d = a_2 - a_1 = 5 \] The sum of the first \( n \) terms of an arithmetic sequence is given by: \[ S_n = \frac{n}{2} [2a_1 + (n - 1)d] \] Substituting \( n = 120 \), \( a_1 = 3 \), and \( d = 5 \): \[ S_{120} = \frac{120}{2} [2(3) + (120 - 1)(5)] = 60 [6 + 595] = 60 \times 601 = 36,\!060 \] **Answer:** The sum of the first 120 terms is **36,060**. --- ### **3.3 The following pattern is true for the arithmetic sequence above:** \[ \begin{array}{l} T_{1}+T_{4}=T_{2}+T_{3} \\ T_{5}+T_{8}=T_{6}+T_{7} \\ T_{9}+T_{12}=T_{10}+T_{11} \\ \vdots \\ \therefore T_{k}+T_{k+3}=T_{x}+T_{y} \end{array} \] #### **3.3.1 Write down the values of \( x \) and \( y \) in terms of \( k \).** Observing the pattern: - When \( k = 1 \): \( x = 2 \), \( y = 3 \) - When \( k = 5 \): \( x = 6 \), \( y = 7 \) - When \( k = 9 \): \( x = 10 \), \( y = 11 \) - ... It becomes clear that: \[ x = k + 1 \quad \text{and} \quad y = k + 2 \] **Answer:** \[ x = k + 1 \quad \text{and} \quad y = k + 2 \] --- #### **3.3.2 Calculate the value of \( T_{x} + T_{y} \) in terms of \( k \) in simplest form.** Given \( x = k + 1 \) and \( y = k + 2 \), and knowing that in an arithmetic sequence: \[ T_n = a_1 + (n - 1)d \] Compute \( T_x \) and \( T_y \): \[ T_x = a_1 + (k + 1 - 1)d = a_1 + k d \] \[ T_y = a_1 + (k + 2 - 1)d = a_1 + (k + 1)d \] Add them together: \[ T_x + T_y = (a_1 + k d) + (a_1 + (k + 1)d) = 2a_1 + (2k + 1)d \] Substitute \( a_1 = 3 \) and \( d = 5 \): \[ T_x + T_y = 2(3) + (2k + 1)(5) = 6 + 10k + 5 = 10k + 11 \] **Answer:** \[ T_{x} + T_{y} = 10k + 11 \] ---