Question

logs are stacked in a pile on the bottom row and 15 on the top row . there are 10 rows in all with each row having one more log that the one above it. how many logs are in the stack?

Answer 1

Since there is a common difference of one, this is an arithmetic sequence of the form a(n)=a+d(n-1) which in this case is:

a(n)=15+1(n-1)

All arithmetic sequences have a sum which is equal to the average of the first and last terms time the number of terms, mathematically:

s(n)=(2an+dn^2-dn)/2  (that is just the result of ((a+a+d(n-1))(n/2)) )

Since a=15 and d=1 this becomes:

s(n)=(30n+n^2-n)/2

s(n)=(n^2+29n)/2 so

s(10)=(100+290)/2

s(10)=195 logs

Answer 2

There are 195 logs in total.

If the top row has 15 logs, and each row below it has one more log than the one above it, then 15 • 10 (since there are 10 rows in total) plus 9+8+7+6+5+4+3+2+1 ( minus the top layer, starting from the 2nd layer down, deducting 1 each time since every time the number gets add to it, a row's requirement that each row below it has one more log gets satisfied.)
(15 • 10) + (9+8+7+6+5+4+3+2+1) = 150 + 45 = 195.