**Q) **250 logs are stacked in the following manner:

22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row?

**Ans: **Let there be n rows to pile of 250 logs

Here, the bottom row has 22 logs and in next row, 1 log reduces

It means, we get an AP 22, 21, 20, 19, ………n with first term or a = 22 and d = -1

Now, we know that total logs are 250 or we can say that,

S_{n} = 250

Since sum of n terms of an A.P. S_{n }= (2a + (n-1) d)

Therefore, 250 = (2 x 22 + (n-1) x (-1))

or 500 = n (44 – (n-1))

500 = n (45 – n)

n^{2} – 45 n + 500 = 0

By solving this, we get (n-20) (n-25) = 0

Since, there are 22 logs in first row and in next row, 1 log reduces, then we can not have more than 22 terms

n ≠ 25

and n = 20

Means, 20^{th} row is the top row of the pile

Now let’s find out number of logs in 20^{th} row

We know that value of n^{th} term of an A.P. = a + (n-1) d

N_{20} = [22 + (20-1) (-1)]

= (22 – 19) = 3

**Therefore, there are 3 logs in the top row.**