Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: RS.$200$ for 1st day, Rs.$250$ for second day, Rs.$300$ for third day and so on. If the contractor pays Rs.$27750$ as penalty, find the number of days for which the construction work is delayed.

Answer 1

Penalty for $1^{st}$ day $=200$

Penalty for $2^{nd}$ day $=250$

Penalty for $3^{rd}$ day $=300$

Here, the series is $200,250,300,...\dots$

Since difference is same, it is A.P.

Common difference $d=50$

We need to find number of days work is delayed.

$\therefore$ We need to find n.

We know that

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$27,750=\frac{n}{2}\left[2\right(200)+(n-1\left).50\right]$

$27,750.2=n[400+50n-50]$

$\Rightarrow 55,500=n[350+50n]$

$\Rightarrow 55,500=350n+50n^2$

$\Rightarrow 50n^2+350n-55,500=0$ $\therefore$ No. of days work is delayed $=30$

$\Rightarrow 50(n^2+7n-1110)=0$

$\Rightarrow n^2+7n-1110=0$

$\Rightarrow n^2+37n-30n-1110=0$

$\Rightarrow n(n+37)-30(n+37)=0$

$\Rightarrow (n+37)(n-30)=0$

$n+37-0$ $n-30=0$

$n=-37$ $n=30$.