Question

200 logs are stacked in such a way that there are 20 logs in a bottom row, 19 logs in the next row, 18 logs in the next row and so on. how many rows are formed and how many logs are there in the top row.

Answer 1

here the series is 20, 19, 18, 17........... hence it becomes an AP.
here common difference(d)= -1
1st term (a)= 20
total i.e Sum. Sn = 200
so applying formula Sn =n/2 (2a +(n-1)d)
200= n/2(2(20+(n-1)(-1))
400= n(41-n)
n^2 - 41n +400=0
so solving we get
n= 16 and n= 25 (two roots for quadratic eqn)
now for n= 16
an = a +(n -1)d
a16 = 20 -(16-1)(-1)
a16 = 20 -15 =5
similarly a25 = 20 -24 =-4
so -4 is not possible for numbers of rows
hence a16 = 5 is the answer
i.e 16 rows and 5 logs

Answer 2

SOLUTION Let the required number of rows be n. Then,

20+ 19 + 18+ .. to n terms = 200.

This is an arithmetic series in which

a = 20, d = (19-20) = -1 and S, = 200.

We know that S 12a + (n-1)d).

2

12 20 + (n − 1)(-1)) = 200

(41 - n) = 400 = n2-41n + 400 = 0

712-25n -16n + 400 = 0 => n(n-25) - 16(1-25) = 0

(n1 - 25) (11-16) = 0 = 11-25 = 0 or 11-16 = 0

n = 25 or n = 16.

Now, T = (a +24d) = 20 + 24 (-1) = -4.

This is meaningless as the number of logs cannot be negative

So, we reject the value n = 25.

11 = 16. Thus, there are 16 rows in the whole stack

Now, T = (a +150) = 20 + 15 X (-1) = 20 - 15 = 5.

Hence, there are 5 logs in the top row.