Arithmetic Progression

General
1

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. In how many rows are the 200 logs placed and how many logs are in the top row? Use LaTex

2

Agrim_Sharma said 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. In how many rows are the 200 logs placed and how many logs are in the top row? Use LaTex

Answer: To solve this problem, we need to use the properties of an arithmetic progression (AP). Given the stacking pattern, we know the logs decrease by 1 log per row, starting from 20. Let’s denote the number of logs in the ( n )-th row by ( a_n ).

Solution By Steps:

  1. Identify the Arithmetic Progression Parameters:

    The first term ( a_1 ) is 20, and the common difference ( d ) is -1 since each subsequent row has 1 less log.

    a_1 = 20, \quad d = -1

  2. Sum of the Arithmetic Progression:

    Let ( S_n ) be the sum of the first ( n ) terms of the AP.

    S_n = \frac{n}{2} \left( 2a_1 + (n-1)d \right)

    We need ( S_n = 200 ).

    \frac{n}{2} \left( 2 \cdot 20 + (n-1) \cdot (-1) \right) = 200

  3. Simplify the Summation Equation:

    \frac{n}{2} \left( 40 - (n-1) \right) = 200
    \frac{n}{2} \left( 41 - n \right) = 200
    n (41 - n) = 400
    n^2 - 41n + 400 = 0

  4. Solve the Quadratic Equation:

    This is a quadratic equation in ( n ). We shall solve it using the quadratic formula ( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ).

    Here, ( a = 1 ), ( b = -41 ), and ( c = 400 ).

    n = \frac{-(-41) \pm \sqrt{(-41)^2 - 4 \cdot 1 \cdot 400}}{2 \cdot 1}
    n = \frac{41 \pm \sqrt{1681 - 1600}}{2}
    n = \frac{41 \pm \sqrt{81}}{2}
    n = \frac{41 \pm 9}{2}

    So, the possible values of ( n ) are:

    n = \frac{50}{2} = 25 \quad \text{or} \quad n = \frac{32}{2} = 16

    We choose ( n = 16 ) because a smaller number of rows for stacking makes practical sense in this context.

  5. Determine the Number of Logs in the Top Row:

    For the 16th row:

    a_{16} = a_1 + (16-1)d
    a_{16} = 20 + 15 \cdot (-1)
    a_{16} = 20 - 15 = 5

Final Answer:

Therefore, the logs are placed in 16 rows, with the top row containing 5 logs.