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

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## 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?

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 19

## 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[ \begin{array} & 2\times 20+ \\ \left( n-1 \right)\left( -1 \right) \end{array} \right] \\ 400&\!=\!n\left[ 40-n+1 \right] \\ 400&\!=\!n\left[ 41-n \right] \\ 400&\!=\!41n-{{n}^{2}} \\ {{n}^{2}}-41n+400&\!=\!0 \\ \left[ \begin{array} & {{n}^{2}}-16n- \\ 25n+400 \\ \end{array} \right]&\!=\!0 \\ \left[ \begin{array} & n\left( n-16 \right)- \\ 25\left( n-16 \right) \\ \end{array} \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 $$5.$$

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