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=\frac{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*15-9\frac{15(15+1)}2`

`S_(15)=-795`

The answer is C.