<s>[INST] Discrete Math Question: arithmetic progression A lumberjack has $4n + 110$ logs in a pile consisting of n layers. Each layer has two more logs than the layer directly above it. If the top layer has six logs, how many layers are there? Write the steps to calculate the equation for the problem and state the number of layers.
It's given that the top layer has 6 logs and each layer below that gains 2 logs per layer.
1st layer: 6
2nd layer: 6 + 1(2) = 6 + 2 = 8
3rd layer: 6 + 2(2) = 6 + 4 = 10
4th layer: 6 + 3(2) = 6 + 6 = 12
5th layer : 6 + 4(2) = 6 + 8 = 14
and so on
the nth layer: $a_n = 6 + 2(n-1)$
1st term 6 and common difference 2 with n terms
1st term: $a_n = 6$
last term: $a_n = 6 + 2n + 4$
$S_n = (n/2)(a_1 + a_n)$
$$4n + 110 = (n/2)(6 + 6 + 2(n-1))$$
Can anyone help break this equation down to solve for n?
 [/INST] As you observe, you have an arithmetic progression, and your work is fine to the point you left off; solving for $n$ is mostly a matter of algebra:
Picking up where you left off 
$$\begin{align} 4n + 110 & = \dfrac n2\left(6 + 6 + 2(n-1)\right) \\ \\
& = \dfrac n2(12) + \dfrac {2n\cdot (n-1)}{2}\\ \\
& = 6n + n(n - 1) \\ \\
& = 6n + n^2 - n \\ \\
4n + 110 & = n^2 + 5n\end{align}$$
Now, we move everything to one side of the equation: $$\begin{align} n^2 + 5n - 4n - 110 & = 0 \\ \\ n^2 + n - 110 & = 0\\ \\(n-10)(n+11) &= 0\end{align}$$
So $n = 10, \; n=-11$ solve the quadratic, but we need the value where $n > 0$. Hence, $n = 10$.
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