# 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.

## 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.