What is the sum of a 44 term arithmetic sequence where the first term is -9 and the last term is 120
Accepted Solution
A:
Answer:2442Step-by-step explanation:The n th term of an arithmetic sequence is[tex]a_{n}[/tex] = a₁ + (n - 1)dwhere a₁ is the first term and d is the common differenceWe require to find d knowing that 120 is the 44 th term, thus[tex]a_{44}[/tex] = - 9 + 43d = 120 ( add 9 to both sides )43d = 129 ( divide both sides by 43 )d = 3The sum to n terms of an arithmetic sequence is[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex] [ 2a₁ + (n - 1)d ], hence[tex]S_{44}[/tex] = [tex]\frac{44}{2}[/tex] [ (2 × - 9 + (43 × 3) ] = 22(- 18 + 129) = 22 × 111 = 2442ORSince the first and last terms in the sequence are known, then[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex] ( first + last )[tex]S_{44}[/tex] = 22(- 9 + 120) = 22 × 111 = 2442