Q:

Find Sn for the given arithmetic series.a1 = 18, d = –3, n = 14

Accepted Solution

A:
The sequence is given recursively by[tex]\begin{cases}a_1=18\\a_n=a_{n-1}-3&\text{for }n>1\end{cases}[/tex]By subsitution, we can solve for [tex]a_n[/tex] in terms of [tex]a_1[/tex]:[tex]a_n=a_{n-1}-3[/tex][tex]a_n=(a_{n-2}-3)-3=a_{n-2}+2(-3)[/tex][tex]a_n=(a_{n-3}-3)+2(-3)=a_{n-3}+3(-3)[/tex]and so on, with[tex]a_n=a_1+(n-1)(-3)\implies a_n=20-3n[/tex]Then the sum of the first 14 terms of the sequence is[tex]S_{14}=\displaystyle\sum_{n=1}^{14}(20-3n)=20\sum_{n=1}^{14}1-3\sum_{n=1}^{14}n[/tex][tex]S_{14}=20\cdot14-\dfrac{3\cdot14\cdot15}2[/tex][tex]\boxed{S_{14}=-35}[/tex]