Question

200 logs are stacked so that 20 logs at bottom row 19 in the next ,18 in the next and so on.In
how many rows can 200 logs be arranged and how many are there on the top row ?

Answer 1

looks like an AS, with a = 20 , d = -1 , n = ?
(n/2)(2a + (n-1)d ) = sum(n)
(n/2)(40 + (n-1)(-1) ) = 200
n( 40 - n + 1) = 400
-n^2 + 41n - 400 = 0
n^2 - 41n + 400 = 0
n = (41 ± √81)/2
= 25 or 16

so term(25) = 20 + 24(-1) = -4 which makes no sense
or
term(16) = 20 + 16(-1) = 4

So there are 16 rows and the last row contains 4 logs