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

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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.


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.


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.