150 workers were engaged to finish a job in a certain number of days.

Let’s assume x to be the number of days in which 150 workers finish the work.

Then from the question, we have

150x = 150 + 146 + 142 + …. (x + 8) terms

The series 150 + 146 + 142 + …. (x + 8) terms is an A.P.

With first term (a) = 150, common difference (d) = –4 and number of terms (n) = (x + 8)

Now, finding the sum of terms:

\( 150x=\dfrac{(x+8)}{2}[2(15)+(x+8-1)(-4)] \)

\( 150x=(x+8)[150+(x+7)(-2)]\)

\( 150x=(x+8)(150-2x-14)\)

\( 150x=(x+8)(136-2x)\)

\( 75x=(x+8)(68-x)\)

\( 75x=68x-x^2+544-8x\)

\( x^2+75x-60x-544=0\)

\( x^2+15x-544=0\)

\( x^2+32x-17x-544=0\)

\( x(x+32)-17(x+32)=0\)

\( (x-17)(x+32)=0\)

x = 17 or x = -32

As x cannot be negative. [Number of days is always a positive quantity]

x = 17

Hence, the number of days in which the work should have been completed is 17.

But, due to the dropping out of workers the number of days in which the work is completed

= (17 + 8) = 25