Question

Find Sn for the given arithmetic series.

a1 = 18, d = –3, n = 14

Answer 1

- 21

Explanation:

The sum to n terms of an arithmetic sequence is

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

`S_(14)` =
`(14)/(2)`[( 2 × 18) + (13 × - 3) ]

= 7 (36 - 39) = 7 × - 3 = - 21

Answer 2

The sequence is given recursively by

`\begin{cases}a_1=18\\a_n=a_(n-1)-3&\text{for }n>1\end{cases}`

By subsitution, we can solve for
`a_n` in terms of
`a_1`:

`a_n=a_(n-1)-3`

`a_n=(a_(n-2)-3)-3=a_(n-2)+2(-3)`

`a_n=(a_(n-3)-3)+2(-3)=a_(n-3)+3(-3)`

and so on, with

`a_n=a_1+(n-1)(-3)\implies a_n=20-3n`

Then the sum of the first 14 terms of the sequence is

`S_(14)=\displaystyle\sum_(n=1)^(14)(20-3n)=20\sum_(n=1)^(14)1-3\sum_(n=1)^(14)n`

`S_(14)=20\cdot14-\frac{3\cdot14\cdot15}2`

`\boxed{S_(14)=-35}`