<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] $$6+8+10+\cdots+(6+2n-2)=6n+(0+2+4+\cdots+2n-2)=\\6n+2(0+1+2+\cdots+n-1)=6n+2\frac{(n-1)n}{2}=6n+(n-1)n=n^2+5n$$
You were told that this equals $4n+110$, so $n^2+5n=4n+110$, which is a quadratic equation $n^2+n-110=(n+11)(n-10)=0$.  The only positive solution is $n=10$.
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