Question

$150$ workers were engaged to finish a job in a certain number of days. $4$ workers dropped out on second day, $4$ more workers dropped out on third day and so on. It took $8$ more days to finish the work. Find the number of days in which the work was completed.

Answer 1

Step 1: Find the total number of workers worked for all $\mathit{n}$ days:

Let the number of days required to complete the work be $n$.
Since $4$ workers dropped on everyday, the number of workers present on successive days are:
$150,146,142,138,\dots \dots .$
If the workers are not dropped, then 150 workers work on everyday to complete the work in $\left(n–8\right)$ days.
The total number of workers worked for all $n$ days will be,

$\Rightarrow S_n=150\left(n–8\right)$

Step 2: Compute the number of days required to complete the work:

Formula:

The sum of $n$ terms in an AP is $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$.
Since the work done in both cases must be same, substitute $S_n=150\left(n–8\right)$, $a=150$ and $a=-4$ into the sum of $n$ terms of an AP formula.
$$\begin{gathered} \Rightarrow 150\left(n-8\right)=\frac{n}{2}[2\times \left(150\right)+\left(n-1\right)\left(-4\right)] \\ \Rightarrow 150n-1200=\frac{n}{2}(304-4n) \\ \Rightarrow 150n-1200=152n-2n^2 \\ \Rightarrow 2n^2-2n-1200=0 \end{gathered}$$

Divide the obtained equation by $2$ and solve for $n$.

$\begin{array}{cc}\Rightarrow & n^2-n-600=0\\ \Rightarrow & n^2-25n+24n-600=0\\ \Rightarrow & n\left(n-25\right)+24\left(n-25\right)=0\\ \Rightarrow & \left(n-25\right)\left(n+24\right)=0\\ \Rightarrow & n=25,-24\end{array}$

Since the value of $n$ cannot be negative, we get $n=25$.

Hence, the work will be completed in $25\mathrm{days}$.