# Ex.5.3 Q19 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

\(200\) 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?

## Text Solution

**What is Known?**

Stack of \(200\) logs, \(20\) logs in the bottom row, \(19\) in the next row, \(18\) in the row next to it and so on.

**What is Unknown?**

Number of rows and number of logs in the top row

**Reasoning:**

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) and \(nth\) term of an AP is\(\,{a_n} = a + \left( {n - 1} \right)d\) Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

It can be observed that the numbers of logs in rows are in 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, \({S_n} = 200\)

We know that sum of \(n\) terms of AP,

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\200 &= \frac{n}{2}\left[ {2 \times 20 + \left( {n - 1} \right)\left( { - 1} \right)} \right]\\400 &= n\left[ {40 - n + 1} \right]\\400 &= n\left[ {41 - n} \right]\\400& = 41n - {n^2}\\{n^2} - 41n + 400& = 0\\{n^2} - 16n - 25n + 400 &= 0\\n\left( {n - 16} \right) - 25\left( {n - 16} \right)& = 0\\\left( {n - 16} \right)\left( {n - 25} \right) &= 0\end{align}\]

Either \(\left( {n - 16} \right) = 0\) or \(\left( {n - 25} \right) = 0\)

\(\therefore n = 16\) or \(n = 25\)

\[\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\{a_{16}}&= 20 + \left( {16 - 1} \right) \times \left( { - 1} \right)\\{a_{16}}& = 20 - 15\\{a_{16}} &= {\rm{ }}5\end{align}\]

Similarly,

\[\begin{align}{a_{25}}&= 20 + \left( {25 - 1} \right) \times \left( { - 1} \right)\\{a_{25}} &= 20 - 24\\{a_{25}} &= - 4\end{align}\]

Clearly, the number of logs in \(16\)th row is \(5.\) However, the number of logs in \(25\rm{th}\) row is negative \(4,\) which is not possible.

Therefore, \(200\) logs can be placed in \(16\) rows.

The number of logs in the top \((16^\rm{th})\) row is \(5.\)