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.
