A construction 2

A construction company will be penalized for its bridge construction delays. The penalty is set at 4,000 for the first day and shall be subjected to a 1,000 increase for each succeeding day. The company can afford a maximum payment of 165,000 for a penalty. Find the maximum number of days by which the completion of work can be delayed.

Correct answer:

n =  15

Step-by-step explanation:

$$\begin{gathered} a_1=4000 \\ d=1000 \\ b=165000 \\ {a}_2=a_1+d=4000+1000=5000 \\ {a}_3=a_2+d=5000+1000=6000 \\ s(n)=\frac{n}{2}[a_1+a_1+(n-1)d] \\ 165000=n|2\cdot (2\cdot a_1+(n-1)\cdot d) \\ 165000=n|2\cdot (2\cdot 4000+(n-1)\cdot 1000) \\ -500n^2-3500n+165000=0 \\ 500n^2+3500n-165000=0 \\ 500=2^2\cdot 5^3 \\ 3500=2^2\cdot 5^3\cdot 7 \\ 165000=2^3\cdot 3\cdot 5^4\cdot 11 \\ \text{GCD}(500,3500,165000)=2^2\cdot 5^3=500 \\ {n}^2+7n-330=0 \\ a=1;b=7;c=-330 \\ D=b^2-4ac=7^2-4\cdot 1\cdot (-330)=1369 \\ D>0 \\ {n}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-7\pm \sqrt{1369}}{2} \\ {n}_{1,2}=\frac{-7\pm 37}{2} \\ {n}_{1,2}=-3.5\pm 18.5 \\ {n}_1=15 \\ {n}_2=-22 \\ n=n_1=15 \end{gathered}$$

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