NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3

Question 19

Step 1: Observe that the rows form an AP.

The number of logs in each row forms an arithmetic progression:

\[20, 19, 18, 17, \ldots\]

So, first term \(a = 20\), common difference \(d = -1\).

Let there be \(n\) rows, and let the last row (top row) have \(l\) logs.

Step 2: Use the sum of AP formula.

Total logs = sum of all terms of the AP:

\[S_n = \frac{n}{2}(a + l)\]

We are given \(S_n = 200\) and \(a = 20\). Also, since \(d = -1\), the nth term is:

\[l = a_n = a + (n - 1)d = 20 + (n - 1)(-1) = 20 - n + 1 = 21 - n\]

So the sum becomes:

\[200 = \frac{n}{2}(20 + 21 - n) = \frac{n}{2}(41 - n)\]

Step 3: Form and solve the quadratic equation.

Multiply both sides by 2:

\[400 = n(41 - n)\]

\[400 = 41n - n^2\]

Rearrange:

\[n^2 - 41n + 400 = 0\]

Factorise: find two numbers whose product is 400 and sum is 41. These are 16 and 25:

\[(n - 16)(n - 25) = 0\]

So, \[n = 16 \quad \text{or} \quad n = 25\]

Step 4: Choose the valid value of \(n\).

The number of logs in the nth row is \(l = 21 - n\). For \(n = 25\):

\[l = 21 - 25 = -4\] (not possible, logs cannot be negative).

So, the only valid solution is \(n = 16\).

Step 5: Find the number of logs in the top row.

Top row logs = nth term:

\[l = 21 - n = 21 - 16 = 5\]

Conclusion: The 200 logs are stacked in 16 rows, and the top row has 5 logs.