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

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Question

200 logs are stacked in such a way that there are 20 logs in the 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?

[Hint: 20 + 19 + 18 + ......... to n terms = 200. Find n and Tn]

Sum
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Solution

The numbers of logs in rows form an AP: 20, 19, 18, ...

Here $$a = 20, \ d = -1$$ and sum = 200. 

Let the number of rows be n.

$$S_n = \frac{n}{2}[2(20) + (n - 1)(-1)] = 200$$

$$\Rightarrow \frac{n}{2}(41 - n) = 200$$

$$\Rightarrow n(41 - n) = 400$$

$$\Rightarrow n^2 - 41n + 400 = 0$$

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

Since the top row cannot have a negative number of logs, $$n = 16$$. 

Top row = $$T_{16} = 20 + 15(-1) = 5$$. 

So there are 16 rows and 5 logs in the top row.

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Chapter 10: Arithmetic Progression - EXERCISE 10B [Page 168]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 10 Arithmetic Progression
EXERCISE 10B | Q 23. | Page 168
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