Question

A construction company will be penalised each day for delay in construction of a bridge. The penalty will be $Rs.4000$ for the first day and will increase by $Rs.1000$ for each following day. Based on its budget, the company can afford to pay a maximum of $Rs.1,65,000$ towards penalty. Find the maximum number of days by which the completion of work can be delayed

Answer 1

Let us write the penalty amount paid by the construction company from the first day as sequence:

$4000,5000,6000,.....$

It is given that the company can pay$165000$ as penalty for this delay at maximum that is $S_n=165000$.

So let us write this amount as arithmetic series as follows:

$4000+5000+6000+.....$

In the above arithmetic series, the first term is $a_1=4000$, second term is $a_2=5000$.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=5000-4000=1000$

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now, substitute $a=4000,d=1000$ and $S_n=165000$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 165000=\frac{n}{2}\left[\right(2\times 4000)+(n-1)1000]\Rightarrow 2\times 165000=n(8000+1000n-1000)\Rightarrow 330000=n(7000+1000n)\Rightarrow 330000=7000n+1000n^2$

$\Rightarrow 1000n^2+7000n-330000=0\Rightarrow 1000(n^2+7n-330)=0\Rightarrow n^2+7n-330=0\Rightarrow n^2+22n-15n-330=0$

$\Rightarrow n(n+22)-15(n+22)=0\Rightarrow (n+22)(n-15)=0\Rightarrow n=-22,n=15$

Ignoring the negative value of $n$ because $n$ represents the number of days delayed, therefore, we get $n=15$.

Hence, the maximum number of days by which the completion of work can be delayed is $15$.