Question

I don't know anything about this and obviously I suck at this lol

Answer 1

An arithmetic sequence is a sequence that starts with some term $a_1$, and the next term is generated by adding a constant $d$ to get the second term $a_2$. Then the next term $a_3$ is generated by adding $d$ to $a_2$, and so on.

Here $a_1=10$ and $d=-9$. So recursively, the sequence is defined by

$\begin{cases}a_1=10\\a_n=a_{n-1}-9&\text{for }n>1\end{cases}$

$S_n$ is the sum of the first $n$ terms of the sequence. Before you can find this sum, an explicit formula for the $n$ would be useful. You have

$a_n=a_{n-1}+d$
$a_n=(a_{n-2}+d)+d=a_{n-2}+2d$
$a_n=(a_{n-3}+d)+2d=a_{n-3}+3d$
and so on, up to
$a_n=a_1+(n-1)d$
(Notice there's a pattern on the right hand side between the subscripted term and the coefficient of $d$. They always add up to $n$.)

So the explicit formula for this sequence is

$a_n=a_1+(n-1)d=10-9(n-1)=19-9n$

You're asked to find the sum of the first 15 terms, which means

$S_{15}=\displaystyle\sum_{n=1}^9a_n=a_1+a_2+\cdots+a_{14}+a_{15}$

Now you could just find the first 9 terms and add them together, but that's more work than necessary. Instead, you have

$S_9=\displaystyle\sum_{n=1}^{15}(19-9n)=19\sum_{n=1}^{15}1-9\sum_{n=1}^{15}n$

There are some well known formulas for the sums of powers:

$\displaystyle\sum_{n=1}^N1=N$
$\displaystyle\sum_{n=1}^Nn=\dfrac{N(N+1)}2$

So you get

$S_{15}=19\displaystyle\sum_{n=1}^{15}1-9\sum_{n=1}^{15}n$
$S_{15}=19\times15-9\dfrac{15(15+1)}2$
$S_{15}=-795$

The answer is C.