3 The sum of the second and sixth terms of an arithmetic sequence is 4 . The third term is 24 more than the \( 11 t^{\mathrm{h}} \) term. 3.1 Determine the first three terms in the sequence. 3.2 Determine a formula for the \( n^{\text {th }} \) term of the sequence. 3.3 Determine the \( 17^{\text {th }} \) term in the sequence. 3.4 Is -72 a term in the sequence? Justify your answer. 4 Which term in the sequence \( -2 ;-\frac{5}{3} ;-\frac{4}{3} ;-1 ; \ldots \) has a value of 8 ? 5 Consider the sequence \( 13 ; 11 ; 9 ; 7 ; 5 ; \ldots \). 5.1 Determine a formula for the \( n^{\text {th }} \) term of the sequence. 5.2 Show that \( \mathrm{T}_{n}+\mathrm{T}_{15-n}=0 \), for \( n \in \mathbb{N}, n<15 \)

**3.1 Determine the first three terms in the sequence.** Let the first term be \( a \) and the common difference be \( d \). - The second term is \( a+d \) and the sixth term is \( a+5d \). The sum is given by \[ (a+d)+(a+5d)=2a+6d=4. \] Dividing by 2, we have \[ a+3d=2. \tag{1} \] - The third term is \( a+2d \) and the \(11^{\mathrm{th}}\) term is \( a+10d \). It is given that \[ a+2d = (a+10d)+24. \] Simplify to obtain \[ a+2d-a-10d=24 \quad\Longrightarrow\quad -8d=24. \] Thus, \[ d=-3. \] - Substitute \( d=-3 \) into equation (1): \[ a+3(-3)=2 \quad\Longrightarrow\quad a-9=2 \quad\Longrightarrow\quad a=11. \] Therefore, the first three terms are: \[ \begin{array}{l} T_1 = 11,\\[1mm] T_2 = 11+(-3)=8,\\[1mm] T_3 = 11+2(-3)=5. \end{array} \] --- **3.2 Determine a formula for the \( n^{\text{th}} \) term of the sequence.** The formula for the \( n^{\text{th}} \) term in an arithmetic sequence is \[ T_n = a + (n-1)d. \] Substitute \( a=11 \) and \( d=-3 \): \[ T_n = 11 + (n-1)(-3)=11-3n+3=14-3n. \] --- **3.3 Determine the \( 17^{\text{th}} \) term in the sequence.** Using the formula \( T_n = 14-3n \) for \( n=17 \): \[ T_{17}=14-3(17)=14-51=-37. \] --- **3.4 Is \(-72\) a term in the sequence? Justify your answer.** To determine whether \(-72\) is a term, we set \[ 14-3n=-72. \] Solve for \( n \): \[ 14-3n=-72 \quad\Longrightarrow\quad -3n=-72-14=-86 \quad\Longrightarrow\quad n=\frac{86}{3}. \] Since \( \frac{86}{3} \) is not an integer, \(-72\) is not a term in the sequence. --- **4. Which term in the sequence \( -2; -\frac{5}{3}; -\frac{4}{3}; -1; \ldots \) has a value of 8?** Let the first term be \( a_1=-2 \) and let the common difference be \( d \). - The difference between the second and first terms is \[ -\frac{5}{3}-(-2)=-\frac{5}{3}+2=\frac{-5+6}{3}=\frac{1}{3}. \] Hence, \( d=\frac{1}{3} \). - The \( n^{\text{th}} \) term is given by \[ T_n = a_1 + (n-1)d = -2+\frac{n-1}{3}. \] Writing with a common denominator: \[ T_n=\frac{-6+n-1}{3}=\frac{n-7}{3}. \] - Set \( T_n=8 \): \[ \frac{n-7}{3}=8 \quad\Longrightarrow\quad n-7=24 \quad\Longrightarrow\quad n=31. \] Thus, the \( 31^{\text{st}} \) term of the sequence is 8. --- **5.1 Determine a formula for the \( n^{\text{th}} \) term of the sequence \( 13; 11; 9; 7; 5; \ldots \).** Let the first term be \( a=13 \) and the common difference be \( d \). - The second term is 11, so \[ d=11-13=-2. \] The \( n^{\text{th}} \) term is given by \[ T_n = a + (n-1)d = 13 + (n-1)(-2)=13-2(n-1). \] Simplify the expression: \[ T_n = 13-2n+2=15-2n. \] --- **5.2 Show that \( T_n+T_{15-n}=0 \), for \( n\in \mathbb{N},\, n<15 \).** Using the formula \( T_n=15-2n \): - The \( n^{\text{th}} \) term is \[ T_n=15-2n. \] - The \((15-n)^{\text{th}}\) term is \[ T_{15-n}=15-2(15-n)=15-30+2n=2n-15. \] Now, add \( T_n \) and \( T_{15-n} \): \[ T_n+T_{15-n}=(15-2n)+(2n-15)=0. \] Thus, it is shown that \[ T_n+T_{15-n}=0 \] for all natural numbers \( n \) such that \( n<15 \).