Consider the following quadratic sequence: \( 87 ; 164 ; 239 ; 312 ; 383 ; \ldots \) 1.1 Write down the next 3 terms. 1.2 Determine the general term. 1.3 Calculate the \( 30^{\text {th }} \) term.

To find the next three terms of the sequence \( 87 ; 164 ; 239 ; 312 ; 383 ; \ldots \), we first calculate the common differences: - \( 164 - 87 = 77 \) - \( 239 - 164 = 75 \) - \( 312 - 239 = 73 \) - \( 383 - 312 = 71 \) The differences decrease by 2 each time. Therefore, the next differences will be \( 69 \), \( 67 \), and \( 65 \). Now, adding these to the last term: - Next term: \( 383 + 69 = 452 \) - Next term: \( 452 + 67 = 519 \) - Next term: \( 519 + 65 = 584 \) So, the next three terms are \( 452, 519, 584 \). To determine the general term, we notice that the second differences of the sequence are constant (they remain at -2): - First differences: \( 77, 75, 73, 71, 69 \) - Second differences: \( -2, -2, -2, -2 \) This indicates that the general term \( T(n) \) can be expressed as a quadratic function. We can write it as: \[ T(n) = an^2 + bn + c \] By forming equations using the given terms: \( T(1) = a(1)^2 + b(1) + c = 87 \) \ \( T(2) = a(2)^2 + b(2) + c = 164 \) \ \( T(3) = a(3)^2 + b(3) + c = 239 \) Solving these equations gives us: 1. \( a + b + c = 87 \) 2. \( 4a + 2b + c = 164 \) 3. \( 9a + 3b + c = 239 \) By subtracting these equations step-by-step, we derive that: - Use equations 1 and 2 to find \( 3a + b = 77 \) - Use equations 2 and 3 to find \( 5a + b = 75 \) Subtract the first from the second: \[ 2a = -2 \implies a = -1 \] Substituting \( a \) back: \[ 3(-1) + b = 77 \implies b = 80 \] And finally, substituting \( a \) and \( b \) back into the first equation gives: \[ -1 + 80 + c = 87 \implies c = 8 \] Thus, the general term is: \[ T(n) = -n^2 + 80n + 8 \] To calculate the \( 30^{th} \) term, plug \( n = 30 \) into the general term: \[ T(30) = -(30)^2 + 80(30) + 8 = -900 + 2400 + 8 = 1508 \] So, the \( 30^{th} \) term is \( 1508 \).