Respuesta :
Answer:
(A)-494
Step-by-step explanation:
Given the arithmetic series
[tex]S_{19}=\sum_{k=1}^{19}4-3k[/tex]
The terms in the sequence are:
- When k=1, 4-3k=4-3(1)=1
- When k=2, 4-3k=4-3(2)=-2
- When k=3, 4-3k=4-3(3)=-5
Therefore, the terms in the sequence are: 1, -2, -5, ...
First term, a =1
Common difference, d=-2-1=-3
The sum of an arithmetic series, [tex]S_n=\dfrac{n}{2}[2a+(n-1)d][/tex]
Therefore:
[tex]S_{19}=\dfrac{19}{2}[2(1)+(19-1)(-3)]\\=9.5[2+18*-3]\\=9.5[2-54]\\=9.5*-52\\=-494[/tex]
The correct option is A.
