QUESTION 23 (NOVEMBER 2014) 23.1 Given the quadratic sequence: \( -1 ;-7 ;-11 ; p ; \ldots \) 23.1.1 Write down the value of \( p \). (2) 23.1.2 Determine the \( n^{\text {th }} \) term of the sequence. (4) 23.1.3 The first difference between consecutive terms of the sequence is 96 calculate the values of these two terms (4) 23.2 The first three terms of a geometric sequence are: \( 16 ; 4 ; 1 \) 23.2.1 Calculate the value of the \( 12^{\text {th }} \) term. (Leave your answer in simplified exponential form.) (3) 23.2.2 Calculate the sum of the first 10 terms of the sequence. (2) 23.3 Determine the value of : \( \quad\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right) \ldots \) up to 98 factors (4)

Let's solve the problem step by step. ### 23.1 Quadratic Sequence The given quadratic sequence is: \( -1, -7, -11, p, \ldots \) #### 23.1.1 Write down the value of \( p \). To find \( p \), we first need to determine the pattern in the sequence. We can calculate the first differences between consecutive terms: 1. First difference between \( -1 \) and \( -7 \): \[ -7 - (-1) = -6 \] 2. First difference between \( -7 \) and \( -11 \): \[ -11 - (-7) = -4 \] The first differences are \( -6 \) and \( -4 \). The second difference (which should be constant for a quadratic sequence) is: \[ -4 - (-6) = 2 \] Since the second difference is constant, we can find the next first difference by adding \( 2 \) to the last first difference: \[ -4 + 2 = -2 \] Now, we can find \( p \): \[ p = -11 + (-2) = -13 \] Thus, the value of \( p \) is \( -13 \). #### 23.1.2 Determine the \( n^{\text{th}} \) term of the sequence. The first differences we have are: - First term: \( -6 \) - Second term: \( -4 \) - Third term: \( -2 \) The first differences form an arithmetic sequence with a common difference of \( 2 \). Let the first term of the quadratic sequence be \( a_1 = -1 \). The first difference \( d_1 = -6 \), the second difference \( d_2 = -4 \), and the third difference \( d_3 = -2 \). The \( n^{\text{th}} \) term can be expressed as: \[ a_n = a_1 + (n-1)d_1 + \frac{(n-1)(n-2)}{2} \cdot d \] where \( d \) is the second difference, which is \( 2 \). Substituting the values: \[ a_n = -1 + (n-1)(-6) + \frac{(n-1)(n-2)}{2} \cdot 2 \] \[ = -1 - 6(n-1) + (n-1)(n-2) \] \[ = -1 - 6n + 6 + n^2 - 3n + 2 \] \[ = n^2 - 9n + 7 \] Thus, the \( n^{\text{th}} \) term of the sequence is: \[ a_n = n^2 - 9n + 7 \] #### 23.1.3 The first difference between consecutive terms of the sequence is 96. Calculate the values of these two terms. Let the two consecutive terms be \( a_k \) and \( a_{k+1} \). The first difference is given by: \[ a_{k+1} - a_k = 96 \] Using the formula for the \( n^{\text{th}} \) term: \[ a_{k+1} = (k+1)^2 - 9(k+1) + 7 \] \[ a_k = k^2 - 9k + 7 \] Now, substituting these into the difference: \[ [(k+1)^2 - 9(k+1) + 7] - [k^2 - 9k + 7] = 96 \] \[ (k^2 + 2k + 1 - 9k - 9 + 7) - (k^2 - 9k + 7) = 96 \] \[ 2k + 1 - 9 = 96 \] \[ 2k - 8 = 96 \] \[ 2k = 104 \] \[ k = 52 \] Now, we can find \( a_{52} \) and \( a_{53} \): \[ a_{52} = 52^2 - 9 \cdot 52 + 7 = 2704 - 468 + 7 = 2243 \] \[ a_{53} = 53^2 - 9 \cdot 53 + 7 = 2809 - 477 + 7 = 2339 \] Thus, the two terms are \( 2243 \) and \( 2339 \). ### 23.2 Geometric Sequence The first three terms of the geometric sequence are: \( 16, 4, 1 \). #### 23.2.1 Calculate the value of the \( 12^{\text{th}} \) term. The common ratio \( r \) can be calculated as: \[ r = \frac{4}{16} = \frac{1}{4} \] The \( n^{\text{th}} \) term of a geometric sequence is given by: \[ a_n = a_1 \cdot r^{n-1} \] where \( a_1 = 16 \) and \( r = \frac{1}{4} \). Thus, the \( 12^{\text{th}} \) term is: \[ a_{12} = 16 \cdot \left(\frac{1}{4}\right)^{11} = 16 \cdot \frac{1}{4^{11}} = 16 \cdot \frac{1}{2^{22}} = \frac{16}{2^{22}} = \frac{2^4}{2^{22}} = \frac{1}{2^{18}} = 2^{-18} \] #### 23.2.2 Calculate the sum of the first 10 terms of the sequence. The sum \( S_n \) of the first \( n \) terms of a geometric sequence is given by: \[ S_n = a_1 \frac{1 - r^n}{1 - r} \] Substituting \( a_1 = 16 \), \( r = \frac{1}{4} \), and \( n = 10 \): \[ S_{10} = 16 \frac{1 - \left(\frac{1}{4}\right)^{10}}{1 - \frac{1}{4}} = 16 \frac{1 - \frac{1}{4^{10}}}{\frac{3}{4}} = 16 \cdot \frac{4}{3} \left(1 - \frac{1}{4^{10}}\right) \] \[ = \frac{64}{3} \left(1 - \frac{1}{4^{10}}\right) \]