A family drills a well. The first metre costs 80,000 dong and every metre after that costs 5,000 dong more than the metre before it. Water is reached at 50 m. What is the total cost?

Step-by-step solution

1. 1

   Identify the arithmetic pattern. The cost of the $k$-th metre is $u_k=80{,}000+(k-1)\cdot 5{,}000$, an AP with $u_1=80{,}000$ and $d=5{,}000$. The total cost is the sum of the first $50$ terms, not $50$ times the first price.

2. 2

   Compute the price of the last metre. $u_{50}=80{,}000+49\cdot 5{,}000=80{,}000+245{,}000=325{,}000\text{ dong}$

3. 3

   Sum the series with the endpoint formula. $S_{50}=\frac{50}{2}\left(u_1+u_{50}\right)=25\,(80{,}000+325{,}000)=25\cdot 405{,}000$

4. 4

   Finish the multiplication. $S_{50}=10{,}125{,}000\text{ dong}$

5. 5

   Check against the distractors. Ignoring the increase gives $50\times 80{,}000=4{,}000{,}000$ (option A), and using $50\times u_{50}$ gives $16{,}250{,}000$. Only the arithmetic-series total $10{,}125{,}000$ (option B) is right, and it equals $50$ metres at the average price of $202{,}500$ dong.