# Ex.5.3 Q15 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: $$\rm{Rs.}\, 200$$ for the first day, $$\rm{Rs.}\, 250$$ for the second day, $$\rm{Rs.}\,300$$ for the third day, etc., the penalty for each succeeding day being $$\rm{Rs.}\,50$$ more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by $$30$$ days.

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 15

## Text Solution

What is Known?

The penalty for delay of completion by a day, $$\rm{Rs} \,200$$ and $$\rm{Rs} \,50$$ more each succeeding day.

What is Unknown?

Amount has to pay as a penalty.

Reasoning:

General form of an arithmetic progression is $$a,{\rm{ }}\left( {a + d} \right),{\rm{ }}\left( {a + 2d} \right),{\rm{ }}\left( {a + 3d} \right).$$

Sum of the first $$n$$ terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Penalty for $$1st$$ day $$\rm{Rs}.\, 200.$$

Penalty for $$2nd$$ day $$\rm{Rs}.\,250$$

Penalty for $$3rd$$ day $$\rm{Rs}.\,300$$

By observation that these penalties are in an A.P. having first term as $$200$$ and common difference as $$50$$ and number of terms as $$30.$$

\begin{align}a &= 200\\d &= 50\\n &= 30\end{align}

Penalty that has to be paid if he has delayed the work by $$30$$ days

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{30}} &= \frac{{30}}{2}\left[ {2 \times 200 + \left( {30 - 1} \right)50} \right]\\ &= 15\left[ {400 + 1450} \right]\\ &= 15 \times 1850\\& = 27750\end{align}

Therefore, the contractor has to pay $$\rm{Rs}. 27750$$ as penalty.

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