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 (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?

**Solution:**

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18…

For the given A.P.,

First term, a = 20 and common difference, d = a_{2 }− a_{1} = 19 − 20 = −1

Let a total of 200 logs be placed in n rows.

Thus, S_{n} = 200

By the sum of the n terms formula,

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

S_{12} = 12/2 [2(20)+( n -1)(-1)]

400 = n (40 − n +1)

400 = n (41 – n )

400 = 41 n − n^{2}

n^{2 }− 41n + 400 = 0

n^{2 }− 16n − 25n +400 = 0

n ( n − 16) − 25( n − 16) = 0

( n − 16)( n − 25) = 0

Either (n −16) = 0 or n − 25 = 0

n = 16 or n = 25

By the nth term formula,

a_{n} = a + ( n − 1) d

a_{16} = 20 + (16 − 1)(−1)

a_{16} = 20 − 15

a_{16} = 5

Similarly, the 25^{th} term could be written as;

a_{25} = 20 + (25 − 1)(−1)

a_{25} = 20 − 24

= − 4

It can be seen, that the number of logs in the 16^{th} row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^{th} row is 5.