QUESTION 3 6; \( 5+x ;-6 ; 6 x \) are the first four terms of a quadratic sequence with \( T n=a n^{2}+b n+c \). 3.1 Show that the value of \( x=-3 \) 3.2 Determine the values of \( a, b \) and \( c \)

To tackle this quadratic sequence, we need to find the second differences, which should be constant. For the sequence 6, \( 5 + x \), -6, and \( 6x \), let's first establish the value of x. By substituting the terms into the sequence, we can solve for \( x \) by ensuring the differences lead to a consistent second difference. If you plug in \( x = -3 \) into the second term, it results in \( 5 + (-3) = 2 \). The sequence becomes 6, 2, -6, and -18. The first differences are -4, -8, and -12, leading to consistent second differences of -4, proving \( x = -3 \). Now for determining a, b, and c, we can use the quadratic formula \( T_n = an^2 + bn + c \). With the four terms 6, 2, -6, and -18 corresponding to \( n = 1, 2, 3, 4 \), we create a system of equations: \( T_1 = a + b + c = 6 \), \( T_2 = 4a + 2b + c = 2 \), and \( T_3 = 9a + 3b + c = -6 \). Solving this system yields a = -2, b = 4, and c = 4. Let’s keep in mind the joy of sequence hunting—there’s a puzzle-like charm to finding patterns in numbers! And once we unveil the constants a, b, and c, we’re equipped to tackle similar problems with ease. Mathematics can be an exciting unveiling game. Keep digging for those numeric treasures!