QUESTION 5 A quadratic pattem has a second term equal to1, a third term equal \( 10-6 \) and a fifth term equal to -14 . 5.1 Calculate the second difference of this quadratic pattern. 5.2 Hence, or otherwise, calculate the first term of the pattern.
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The second difference of a quadratic sequence is constant and can be found by subtracting the first differences of the sequence. In this case, if we denote the terms of the quadratic sequence as \( a_1, a_2, a_3, a_4, ... \), we know \( a_2 = 1 \), \( a_3 = 10 - 6 = 4 \), and \( a_5 = -14 \). To find the second difference, we can determine the first differences: \( a_3 - a_2 = 4 - 1 = 3 \) and \( a_5 - a_4 \) will need \( a_4 \) first, which we can find from patterns. To obtain \( a_4 \), we can try different values based on our known terms, but if you explore how they relate, the second difference \(\Delta^2\) would ultimately be consistent, and from the pattern, we expect it to be \( -8 \). The first term can be calculated using the formula for a quadratic sequence, knowing that \( a_n = a + (n-1)d + \frac{(n-1)(n-2)}{2}c \), where \( c \) is the second difference. Plugging our values into this formula for \( n=1 \) allows us to solve for \( a_1 \). Given the consistent second difference, we can form equations to arrive at \( a_1 = 3 \).