Question

What is the sum of a 44 term arithmetic sequence where the first term is -9 and the last term is 120

Answer 1

2442

Explanation:

The n th term of an arithmetic sequence is

`a_(n)` = a₁ + (n - 1)d

where a₁ is the first term and d is the common difference

We require to find d knowing that 120 is the 44 th term, thus

`a_(44)` = - 9 + 43d = 120 ( add 9 to both sides )

43d = 129 ( divide both sides by 43 )

d = 3

The sum to n terms of an arithmetic sequence is

`S_(n)` =
`(n)/(2)` [ 2a₁ + (n - 1)d ], hence

`S_(44)` =
`(44)/(2)` [ (2 × - 9 + (43 × 3) ]

= 22(- 18 + 129)

= 22 × 111 = 2442

OR

Since the first and last terms in the sequence are known, then

`S_(n)` =
`(n)/(2)` ( first + last )

`S_(44)` = 22(- 9 + 120) = 22 × 111 = 2442