Case Study III

200 logs are stacked in the following manner:
20 logs in the bottom row, 19 in the next row, 18 in the next row and so on.

Based on this information, answer the following questions:

1. In how many rows these 200 logs are placed?
   (a) 25
   (b) 20
   (c) 16
   (d) 14

2. The number of logs in the top row is:
   (a) 1
   (b) 5
   (c) 3
   (d) 8

3. The number of logs in the 8th row from the bottom is:
   (a) 14
   (b) 11
   (c) 12
   (d) 13

4. Total number of logs in the first six rows from the bottom is:
   (a) 105
   (b) 95
   (c) 85
   (d) 75

5. The number of logs in the 5th row from the top is:
   (a) 8
   (b) 9
   (c) 7
   (d) 10

1. Given,

Logs stacked in rows form an A.P.: 20, 19, 18…..

S_n = 200

a = 20

d = 19 - 20 = -1

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Let 200 logs be stacked in n rows.

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

⇒ 200 × 2 = 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 - 25)(n - 16) = 0

⇒ (n - 25) = 0 or (n - 16) = 0

⇒ n = 25 or n = 16.

If n = 25, a₂₅ = a + 24d = 20 + 24(-1) = -4 (Impossible, the 25th row would have –4 logs).

If n = 16, a₁₆ = a + 15d = 20 + 15(-1) = 5 (Valid).

The number of rows is 16.

Hence, option (c) is the correct option.

2. Given,

Logs stacked in rows form an A.P.: 20, 19, 18, …..

a = 20

n = 16 (16 total rows of logs in the stack i.e top row is 16th.)

d = -1

We know that,

a_n = a + (n - 1)d

⇒ a₁₆ = 20 + (16 - 1)(-1)

= 20 + 15(-1)

= 20 - 15

= 5.

The number of logs in the top row is 5.

Hence, option (b) is the correct option.

3. Given,

Logs stacked in rows form an A.P.: 20, 19, 18…..

a = 20

Total rows = 16

n = 8, (A.P. starts from bottom only)

d = -1

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₈ = 20 + (8 - 1)(-1)

= 20 + 7(-1)

= 20 - 7

= 13.

The number of logs in the 8th row from the bottom is 13.

Hence, option (d) is the correct option.

4. Given,

Logs stacked in rows form an A.P.: 20, 19, 18, …..

a = 20

d = -1

n = 6

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₆ = $\dfrac{6}{2}$ [2(20) + (6 - 1)(-1)]

= 3(40 - 5)

= 3 × (35)

= 105.

Total number of logs in the first six rows from the bottom is 105.

Hence, option (a) is the correct option.

5. The 5th row from the top is the (16 - 5 + 1) = 12th row from the bottom (a₁₂):

Logs stacked in rows form an A.P.: 20, 19, 18…..

We know that,

a_n = a + (n - 1)d

⇒ a₁₂ = 20 + (12 - 1)(-1)

= 20 - 11

= 9.

Hence, option (b) is the correct option.