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]
Answer:
- 21
Step-by-step explanation:
The sum to n terms of an arithmetic sequence is
[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex] [ 2a₁ + (n - 1)d ], thus
[tex]S_{14}[/tex] = [tex]\frac{14}{2}[/tex][( 2 × 18) + (13 × - 3) ]
= 7 (36 - 39) = 7 × - 3 = - 21
