Let the no. of days the work was to be completed be $n$
Given: No. of workers on $1^{st}$ day $=150$
Each day $4$ workers dropped on the subsequent days.
Also given that it took $8$ more days to finish the work.
$\therefore$ Total number of days the work went on $=n+8$
Step 1
If 150 men work for $n$ days the work will be finished by $n$ days.
$\therefore$ The work done by $1$ man on one day $ =\large\frac{1}{n\times 150}$
On first day no. of workers = 150$
$\therefore$ The work done on first day $=\large\frac{1}{n\times 150}$$\times 150$
Step 2
the no. of workers on second day $= 146$
$\therefore$ The work done on second day $=\large\frac{1}{n\times 150}$$\times 146$
Similarly the no. of workers on third day $142$
$\therefore$ The work done on third day $=\large\frac{1}{n\times 150}$$\times 142$
and so on
the work goes on for $n+8$ days
Step 3
Thus the if we assume the total work to be completed as $1$, then
$1=\large\frac{1}{n\times 150}$$\times 150$$+\large\frac{1}{n\times 150}$$\times 146$$+\large\frac{1}{n\times 150}$$\times 142$$+........$(upto $n+8$ terms.)
Taking $\large\frac{1}{150n}$ common we get
$1=\large\frac{1}{150 n}$$\big(150+146+142+.......$ upto $(n+8)$ terms$\big)$
$\big(150+146+142+.......$ upto $(n+8)$ terms$\big)$ is an A.P.
with first term $a=150$ and common difference $d=-4$
We know that sum of $n$ terms of an A.P. $=\large\frac{n}{2}$$\big[2a+(n-1)d\big]$
$\Rightarrow\:150n=\large\frac{n+8}{2}$$\big[2\times 150+(n+8-1)(-4)\big]$
$\Rightarrow\:150n=\large\frac{n+8}{2}$$\big[300-(n+7)(4)\big]$
$\Rightarrow\:150n=\large\frac{n+8}{2}$$\big[300-4n-28\big]$
$\Rightarrow\:150n=\large\frac{n+8}{2}$$\big[272-4n\big]$
$\Rightarrow\:150n=(n+8)(-2n+136)$
$\Rightarrow\:2n^2+30n-1088=0$
$\Rightarrow\:n^2+15n-544=0$
$\Rightarrow\:n^2+32n-17n-544=0$
$\Rightarrow\:(n-17)(n(+32)=0$
$\Rightarrow\:n=17\:\:or\:\:-32$
But since $n$ cannot be $-ve$, $n=17$
$\therefore$ The no. of days the work was completed $=17+8=25\:days.$