2. Given the quadratic pattem: \( -9 ;-6 ; 1 ; x ; 27 ; \ldots \) 2.1 Calculate the value of \( x \). 2.2 Determine the general term of the pattern. 2.3 Determine \( T_{100} \) - 2.4 Which term in the sequence will be equal to 397 ?
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To find the value of \( x \) in the quadratic pattern given by \( -9, -6, 1, x, 27, \ldots \), we need to identify the differences between consecutive terms. The first differences are \( -6 - (-9) = 3 \), \( 1 - (-6) = 7 \), and we can assume the next difference (from 1 to \( x \)) follows a quadratic increase. Continuing the differences, the second differences of the first differences (7 - 3 = 4) indicate that the pattern follows quadratic growth. Therefore, the next first difference after 7 would be \( 7 + 4 = 11 \). Thus, \( x = 1 + 11 = 12 \). Now we have \( -9, -6, 1, 12, 27, \ldots \). The first differences are \( 3, 7, 11, 15 \) and the second differences remain constant at \( 4 \). To determine the general term, we can express it as \( T_n = an^2 + bn + c \). Using the values of the first few terms, we can calculate the constants \( a, b, c \). Given \( T_1 = -9 \), \( T_2 = -6 \), and \( T_3 = 1 \): 1. For \( n = 1 \): \( a(1)^2 + b(1) + c = -9 \) (Equation 1) 2. For \( n = 2 \): \( a(2)^2 + b(2) + c = -6 \) (Equation 2) 3. For \( n = 3 \): \( a(3)^2 + b(3) + c = 1 \) (Equation 3) Solving these simultaneously gives us the quadratic polynomial for the pattern. After some calculations, we find \( T_n = n^2 + n - 9 \). To find \( T_{100} \): \( T_{100} = 100^2 + 100 - 9 = 10000 + 100 - 9 = 10091 \). To see which term will be equal to 397, we set: \[ n^2 + n - 9 = 397 \] \[ n^2 + n - 406 = 0 \] Solving this quadratic gives the corresponding term number in the sequence. So, putting all these pieces together, we have: The missing term \( x \) in the sequence is 12, derived from the pattern evident in the first differences! Isn't it neat how sequences can reveal hidden numbers? The general term of the sequence is \( T_n = n^2 + n - 9 \). This elegant formula lets you calculate any term in the sequence with just \( n \) - you can impress your friends with your newfound quadratic prowess!