N.B. \( \mathrm{d}=\mathrm{T}_{2}-\mathrm{T}_{1}=\mathrm{T}_{3}-\mathrm{T}_{2} \) General Term: \( \mathrm{T}_{\mathrm{n}}=\frac{a+(n-1) d}{} \) where \( d=T 2-T 1 \) Sum of n Terms: \( \mathrm{S}_{\mathrm{n}}=\frac{n}{2}\left(2 a+(n-1) d\right. \) or \( \frac{n}{2}(a+l) \) where \( l= \) last term 1. Given the arithmetic series: \( 2+5+8+\ldots+62 \) Calculate 1.1 The number of terms in the series 1.2 The sum of this series 2. The following is an arithmetic sequence: \( 1-\mathrm{p} ; 2 \mathrm{p}-3 ; \mathrm{p}+5 ; \ldots \) 2.1 Calculate the value of p 2.2 Write down the value of: 2.2.1 The first term of the sequence 2.2.2 The common difference 2.3 Explain why none of the numbers in this arithmetic sequence are perfect squar 3. Consider the arithmetic sequence: \( -8 ;-2 ; 4 ; 10 ; \ldots \) 3.1 Write down the next term of the sequence 3.2 If the nth term of the sequence is 148, determine the value of \( n \)

Sure, let's tackle each problem step by step. --- ## **Problem 1** **Given the arithmetic series:** \( 2 + 5 + 8 + \ldots + 62 \) ### **1.1. Calculate the number of terms in the series** **Given:** - First term (\( a \)) = 2 - Common difference (\( d \)) = \( 5 - 2 = 3 \) - Last term (\( l \)) = 62 **Formula for the \( n \)-th term of an Arithmetic Sequence:** \[ T_n = a + (n - 1) \cdot d \] **Plugging in the known values:** \[ 62 = 2 + (n - 1) \cdot 3 \] **Solving for \( n \):** \[ 62 - 2 = 3(n - 1) \\ 60 = 3(n - 1) \\ n - 1 = \frac{60}{3} \\ n - 1 = 20 \\ n = 21 \] **Answer:** There are **21 terms** in the series. ### **1.2. Calculate the sum of this series** **Formula for the sum of the first \( n \) terms of an Arithmetic Series:** \[ S_n = \frac{n}{2} \cdot (a + l) \] **Plugging in the known values:** \[ S_{21} = \frac{21}{2} \cdot (2 + 62) = \frac{21}{2} \cdot 64 = 21 \cdot 32 = 672 \] **Answer:** The sum of the series is **672**. --- ## **Problem 2** **Given the arithmetic sequence:** \( 1 - p ;\ 2p - 3 ;\ p + 5 ; \ldots \) ### **2.1. Calculate the value of \( p \)** **Understanding the sequence:** - First term (\( T_1 \)) = \( 1 - p \) - Second term (\( T_2 \)) = \( 2p - 3 \) - Third term (\( T_3 \)) = \( p + 5 \) **Common difference (\( d \)) should be constant:** \[ d = T_2 - T_1 = (2p - 3) - (1 - p) = 3p - 4 \] \[ d = T_3 - T_2 = (p + 5) - (2p - 3) = -p + 8 \] **Setting the two expressions for \( d \) equal:** \[ 3p - 4 = -p + 8 \] \[ 3p + p = 8 + 4 \\ 4p = 12 \\ p = 3 \] **Answer:** \( p = 3 \) ### **2.2. Write down the value of:** **With \( p = 3 \), let's find the specific values.** #### **2.2.1. The first term of the sequence** \[ T_1 = 1 - p = 1 - 3 = -2 \] **Answer:** The first term is **-2**. #### **2.2.2. The common difference** \[ d = 3p - 4 = 3 \cdot 3 - 4 = 9 - 4 = 5 \] **Answer:** The common difference is **5**. ### **2.3. Explain why none of the numbers in this arithmetic sequence are perfect squares** **Sequence with \( p = 3 \):** \( T_n = -2, 3, 8, 13, 18, \ldots \) **General term:** \[ T_n = a + (n - 1)d = -2 + 5(n - 1) = 5n - 7 \] **Assume \( T_n \) is a perfect square:** \[ 5n - 7 = k^2 \quad \text{for some integer } k \] **Rearrange:** \[ 5n = k^2 + 7 \\ k^2 \equiv -7 \pmod{5} \\ k^2 \equiv 3 \pmod{5} \] **Checking quadratic residues modulo 5:** - \( 0^2 \equiv 0 \) - \( 1^2 \equiv 1 \) - \( 2^2 \equiv 4 \) - \( 3^2 \equiv 4 \) - \( 4^2 \equiv 1 \) **Possible residues:** 0, 1, 4 **Since 3 is not a quadratic residue modulo 5, there exists no integer \( k \) such that \( k^2 \equiv 3 \pmod{5} \).** **Conclusion:** There are **no perfect squares** in the sequence. --- ## **Problem 3** **Given the arithmetic sequence:** \( -8 ; -2 ; 4 ; 10 ; \ldots \) ### **3.1. Write down the next term of the sequence** **Identify the common difference (\( d \)):** \[ d = T_2 - T_1 = -2 - (-8) = 6 \\ d = T_3 - T_2 = 4 - (-2) = 6 \\ d = T_4 - T_3 = 10 - 4 = 6 \] **Next term (\( T_5 \)):** \[ T_5 = T_4 + d = 10 + 6 = 16 \] **Answer:** The next term is **16**. ### **3.2. If the \( n \)-th term of the sequence is 148, determine the value of \( n \)** **Given:** \[ T_n = 148 \] **Formula for the \( n \)-th term:** \[ T_n = a + (n - 1)d \] **Plugging in the known values:** \[ 148 = -8 + (n - 1) \cdot 6 \] \[ 148 + 8 = 6(n - 1) \\ 156 = 6(n - 1) \\ n - 1 = \frac{156}{6} = 26 \\ n = 27 \] **Answer:** \( n = 27 \) ---