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marv_1
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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.
 one year ago
 one year ago
marv_1 Group Title
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.
 one year ago
 one year ago

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jessica.17 Group TitleBest ResponseYou've already chosen the best response.0
So for the topmost layer you have 6 logs, then you have 8, then 10 etc. 6 + (6+2) + (6+2+2)+...+(6+2(n1)) = 4n + 110. You have to find the sum of an arithmetic sequence with first term 6 and common difference 2 with n terms. First term: 6, Last term: 2n+4. Then substitute Sum of sequence = 4n+110 and solve for n.
 one year ago

marv_1 Group TitleBest ResponseYou've already chosen the best response.0
okay this would be 4 divided by 110?
 one year ago

jessica.17 Group TitleBest ResponseYou've already chosen the best response.0
Sorry, what is 25n? Sum of sequence or?
 one year ago

marv_1 Group TitleBest ResponseYou've already chosen the best response.0
no the answer
 one year ago

marv_1 Group TitleBest ResponseYou've already chosen the best response.0
I'm really not sure how to do this
 one year ago

jessica.17 Group TitleBest ResponseYou've already chosen the best response.0
Okay, so how would you find the sum of an arithmetic series? Say, if I told you the first term was 6, the last was 14, and the common difference was 2. (6 + 8 + 10 + 12 + 14.) Apart from adding them up one by one.
 one year ago

jessica.17 Group TitleBest ResponseYou've already chosen the best response.0
Since the terms increased by the same amount each time, you could take the average and multiply it by the number of terms. The average would be (first term + last term)/2.
 one year ago
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