Here's my question:
For the series shown, find n such that $S_{n} = -4290$:
42 + 34 + 26 + 18 + …
I don't want a direct answer to the question, just a way to determine $A_{n}$ so I can do the work on my own.
Edit: $n$ and $A_{n}$ are unknown at the moment.
On
You have $a_n=42-8(n-1)$. Now use the sum formula $$\begin {align}-4290&=\frac 12n(a_1+a_n)\\&=\frac 12n(42+(42-8(n-1))\\ &=\frac 12n(84-8n+8)\\&=46n-4n^2\\ &4n^2-46n-4290=0\end {align}$$ and solve the quadratic for $n$
On
It isn't hard to figure out that $A_{n}=42-8\cdot (n-1)$, then $$A_{1}=42$$ $$A_{2}=34$$ $$A_{3}=26$$ $$A_{4}=18$$ and so on...
$$S_{n}=\sum_{k=1}^{n}A_{k}=\sum_{k=1}^{n}(42-8\cdot (k-1))=\sum_{k=1}^{n}42-8\sum_{k=1}^{n}(k-1)=42n-8(\sum_{k=1}^{n}k-\sum_{k=1}^{n}1)=42n-8(\sum_{k=1}^{n}k-n)$$ Since we know that $\sum_{k=1}^{n}k=\text{ [sum of all natural numbers not greater than n] } = \frac{n\cdot (n+1)}{2}$, we can simplify: $$S_{n}=42n-8\cdot \left(\frac{n\cdot (n+1)}{2}-n\right)=42n-4n\cdot(n+1)+8n=50n-4n\cdot(n+1)=50n-4n^{2}-4n=\color{red}{46n-4n^{2}}$$
Hints:
If you can find a formula $f(n)$ for the sum of the first $n$ terms, then you have only to solve $f(n)=-4290$ for $n$.
If the first term is $a$, and the difference between successive terms is $d$, then the terms are $a$, $a+d$, $a+2d$, $a+3d$, and so on.
The sum with one term is $a =\boxed{1a}$.
The sum with two terms is $a+ \underbrace{(a+d)}_{2\textrm{nd term}}=\boxed{2a+(1)d}$.
The sum with three terms is $\underbrace{2a+(1)d}_{\textrm{sum of }2\textrm{ terms}} + \underbrace{(a+2d)}_{3\textrm{rd term}}=\boxed{3a + (1+2)d}$.
The sum with four terms is $\underbrace{3a + (1+2)d}_{\textrm{sum of }3\textrm{ terms}} + \underbrace{(a+3d)}_{4\textrm{th term}}=\boxed{4a + (1+2+3)d}$.
The sum with five terms is $\underbrace{4a + (1+2+3)d}_{\textrm{sum of }4\textrm{ terms}} + \underbrace{(a+4d)}_{5\textrm{th term}}=\boxed{5a + (1+2+3+4)d}$.
So you need to ask yourself: