Question

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​

Answer 1

Answer:

there are 16 rows and 5 in the top row

Step-by-step explanation:

s_{n} = n/2 [2a + (n-1) d]\\

200 = n/2 [ 2 x 20 + (n-1)(-1)]\\

400 =n [ 40- n +1 ]\\

400 =n[ 41 -n]\\

400 = 41n -n^{2} \\

n^{2} -41n +400 =0\\

n^{2} - 16n -25n +400 =0\\

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

n=16 or 25

if n=25 ,

an =20 +24(-1)

an =20-4

an= -4

if n=16,

an= 20 +15( -1)

an =20-15

an=5

hence we ignore n = 25 as the no of logs in a negative term is impossible

therefore 16 rows and it consists of 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.