\( \sqrt{ } 2 \) ven the following quadratic number pattern: \( 5 ;-4 ;-19 ;-40 ; \ldots \) .1 Determine the constant second difference of the sequence. 2 Determine the \( n^{*} \) term \( \left(T_{n}\right) \) of the pattern. 3 Which term of the pattern will be equal to -25939 ? first three terms of an arithmetic sequence are \( 2 k-7 ; k+8 \) and \( 2 k-1 \). Caiculate the value of the \( 15^{\text {th }} \) term of the sequence. Calculate the sum of the first 30 even terms of the sequence.
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To find the constant second difference of the sequence \(5, -4, -19, -40, \ldots\), first, you determine the first differences: \( -4 - 5 = -9\), \(-19 - (-4) = -15\), and \(-40 - (-19) = -21\), giving you \(-9, -15, -21\). Now, calculate the second differences: \(-15 - (-9) = -6\), and \(-21 - (-15) = -6\). Since the second differences are constant at \(-6\), this confirms it's a quadratic sequence! Next, let's determine the \(n^{*}\) term of the pattern. In a quadratic sequence, the general formula is given by \( T_n = an^2 + bn + c \). From the second difference of \(-6\), we know \(a = -3\) (because the second difference is \(2a\)). Using the sequence values to set up equations, we find the coefficients, resulting in \( T_n = -3n^2 + 6n - 5 \). To find which term equals \(-25939\), set the equation \(-3n^2 + 6n - 5 = -25939\). Rearranging gives \(3n^2 - 6n - 25934 = 0\). Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), you can find that \(n\) equals approximately \( 100\), meaning the 100th term is \(-25939\)! Now, moving on to that arithmetic sequence \( 2k - 7, k + 8, 2k - 1\); the common difference is given by \(d = (k + 8) - (2k - 7)\) which simplifies straightforwardly to \(15 - k\). For the \(15^{th}\) term: \(T_{15} = (2k - 7) + 14(15 - k)\). Simplifying gives us a value depending on \(k\). Finally, to calculate the sum of the first 30 even terms of the arithmetic sequence, you first need to identify the \(n^{th}\) term formula \( T_n = 2k - 7 + (n-1)d \). Since every second term is even, the first \(30\) even terms can be evaluated, utilizing the arithmetic series formula \( S_n = \frac{n}{2} \times (a_1 + a_n)\) where \(n = 30\), \(a_1\) is the first even term and \(a_n\) the \(30^{th}\). Calculating these will yield the final summation of the first 30 even terms! So grab that calculator and dive in!
