Question

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

Answer 1

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

$S_{14}=\displaystyle\sum_{n=1}^{14}(20-3n)=20\sum_{n=1}^{14}1-3\sum_{n=1}^{14}n$

$S_{14}=20\cdot14-\dfrac{3\cdot14\cdot15}2$

$\boxed{S_{14}=-35}$

Answer 2

Answer:

- 21

Step-by-step explanation:

The sum to n terms of an arithmetic sequence is

$S_{n}$ = $\frac{n}{2}$ [ 2a₁ + (n - 1)d ], thus

$S_{14}$ = $\frac{14}{2}$[( 2 × 18) + (13 × - 3) ]

= 7 (36 - 39) = 7 × - 3 = - 21