200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Question

Philoid · Accepted Answer

Let us consider the row number as the term in the series & the no. Of logs present in the each row as the values

20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on.

Considering bottom row as first term so (a) = 20

Second term a₂ = 19

Common difference (d) = a₂ – a = 19-20 = -1

Given that 200 logs are stacked

So, s_n = 200

S_n = [2a + (n-1) d]

Here n is the no. Of rows are used to pile the log.

S_n = [2×20+ (n-1)(-1)]

200 = (41-n)

400 = 41n – n²

n² -41n+400 = 0

n² – 25n -16n + 400 = 0

n(n-25) -16(n-25) = 0

(n-16)(n-25) = 0

∴ n = 16 or 25

Let us take n = 16, because 25 is higher value and that is not feasible solution for the problem

a₁₆ = a+(16-1)d = 20+15×(-1) = 5

There will be 5 logs in top row.