Stones

Ok then consider a row of $2n + 1$ stones where $A$ is the srone in the middle and there are $n$ stones on the left and right of $A$ . Now it is told that he starts from one end of the stones. Suppose he started from the right. Then first he will take the first stone of the RHS bring it to A and put it there. Then he will take the 2nd stone and bring it to A and drop it. He continues this till he collects all the stones on the right hand side of $A$ . So, in doing this he covered a distance of twice the sum of the terms of the AP $10, 20 , 30 , 40, \cdots , 10(n - 1) , 10n$ but he has taken the first stone only once towards A. So, we have to subtract 10n from this. So, the total distance covered in carrying all the stones from the RHS is $10n + 2\left(10 + 20 + 30 + 40 + \cdots + 10(n - 1)\right)$ .

Now currently he is at A. Now he begins to collect all the stones on the LHS. He takes the first stone and come back to A, then the second one and come back to A, then the third, then the fourth and atlast the nth one. So, the total distance covered by him is twice the sum of the terms of the AP $10, 20, 30, 40, \cdots 10n$ . Nothing is subtracted here since he starts and ends at the same place i.e. A. So, the distance covered by him by collecting all stones of the LHS = $2\left(10 + 20 + 30 + 40 + \cdots + 10n\right)$ .

It is given that the total distance travelled is $3km = 3000m$ .

So,
$10n + 2\left(10 + 20 + 30 + 40 + \cdots + 10(n - 1)\right) + 2\left(10 + 20 + 30 + 40 + \cdots + 10n\right) = 3000\\ 10n + 2\left(10 + 20 + 30 + 40 + \cdots + 10(n - 1)\right) + 2×10n + 2\left(10 + 20 + 30 + 40 + \cdots + 10(n - 1)\right) = 3000\\ 30n + 4\left(10 + 20 + 30 + 40 + \cdots + 10(n - 1)\right) = 3000\\ 30n + 4\left(\dfrac{n-1}{2}\left(10 + 10\left(n-1\right)\right)\right) = 3000\\ 30n + 4\left(\dfrac{n-1}{2}\left(10n\right)\right) = 3000\\ 30n + 20n\left(n - 1\right) = 3000\\ 30n + 20n^2 - 20n = 3000\\ 20n^2 + 10n - 3000 = 0\\ 10\left(2n^2 + n - 300\right) = 0\\ 2n^2 + n - 300 = 0\\ 2n^2 - 24n + 25n - 300 = 0\\ 2n\left(n - 12\right) + 25n\left(n - 12\right) = 0\\ \left(n - 12\right)\left(2n + 25\right) = 0\\ n - 12 = 0 \left(\text{or}\right) 2n + 25 = 0\\ n = 12 \left(\text{or}\right) n = \dfrac{-25}{2}$

But number of stones cannot be negative, so, $n \neq \dfrac{-25}{2}$ . So, $n = 12$ .

$\therefore$ Number of stones = $\left(2n + 1\right) = \left(2×12 + 1\right) = \color{#69047E}{\boxed{25}}$ .