# 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:**

The sum of the first n terms of an Arithmetic Progression is given by Sₙ = n/2 [2a + (n - 1) d].

Here, a is the first term, d is a common difference and n is the number of terms.

It can be observed that the number of logs in rows are forming an A.P. 20, 19, 18, ...

For this A.P.,

- First terms, a = 20
- Common difference, d = 19 - 20 = - 1
- Sum of the n terms, Sₙ = 200

We know that sum of n terms of AP is given by the formula Sₙ = n/2 [2a + (n - 1) d]

200 = n/2 [2 × 20 + (n - 1)(- 1)]

400 = n [40 - n + 1]

400 = n [41 - n]

400 = 41n - n²

n² - 41n + 400 = 0

n² - 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

The number of logs in nth row will be,

aₙ = a + (n - 1) d

a₁₆ = 20 + (16 -1) × (- 1)

a₁₆ = 20 - 15

a₁₆ = 5

Similarly,

a₂₅ = 20 + (25 - 1) × (- 1)

a₂₅ = 20 - 24

a₂₅ = - 4

Clearly, the number of logs in the 16th row is 5. However, the number of logs in the 25th row is negative 4, which is not possible.

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

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 5

**Video Solution:**

## 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?

Class 10 Maths NCERT Solutions& Chapter 5 Exercise 5.3 Question 19

**Summary:**

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 then 200 logs can be placed in 16 rows. The number of logs in the top row is 15.

**☛ Related Questions:**

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- How many terms of the AP. 9, 17, 25 ... must be taken to give a sum of 636?
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