Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: $Rs.200$ for the first day, $Rs.250$ for the second day, $Rs.300$ for the third day, etc., the penalty for each succeeding day being $Rs.50$ 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.

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Solution

As it has been given in the question that the penalty for each succeeding day being $Rs.50$ more than for the preceding day. This condition is in the form of an Arithematic progression (A.P.) Such that

$200,250,300,350,400,andsoon$

Where $\mathrm{a}=200\mathrm{and}\mathrm{d}=50$.

Now, we need to find the penalty if the work is delayed by $30$ days. Therefore, $n=30$

We know that the sum of an A.P.

${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{30}=\frac{30}{2}\left[2\times 200+\left(30-1\right)50\right]\phantom{\rule{0ex}{0ex}}{S}_{30}=15\times 1850\phantom{\rule{0ex}{0ex}}{S}_{30}=27750$

**Hence, the total penalty given by the contractor is **$\u20b927750$.

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