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?

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

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.