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There are odd number of rocks aligned in a row. Distance between each rock 10 meters. These rocks should be collected at the middle stone. A worker can carry only one rock at the same time. After collecting all these rocks, he walked total of 3 kilometers. How many rocks are there?

I approached this by only considering one side of middle stone, because the same principle would apply to the other side. Because the distance between each rock is 10 meters, to collect the first rock he has to travel $10*2$, first to get the rock and the then to make his way back to the middle stone. For the second rock, he has to travel $20*2$, for the third $30*2$ and so on. Sum of all these distances traveled has to be equal to $1.5km=1500 m$, because we are only considering half of the stones.

So this is arithmetic series where we have: $a_1=20$, $d=20$ where $a$ is the first term and $d$ is common difference. By calculating arithmetic sum:
$1500=frac{2*20+20(n-1)}{2}n$ where $n$ is number of rocks on one side of the row

Thus $n=-frac{1}{2}±frac{sqrt{601}}{2}$, which is clearly not true, because number of rocks has to be an integer.

Where am I making a mistake? I appreciate any help.

There are $25$ rocks.

The worker starts at the first rock (from the left) and carries that rock to rock 13, which is the middle rock. He then goes back to get the second rock and brings it to the middle rock and so on. When all rocks to the left are collected, he collects the ones on the right.

The distance covered carrying the rocks on the left side is then $$D_L = 120 + 20(1+2+3+ldots+11) = 1,440 m$$
The distance covered carrying the rocks on the right side is $$D_R = 20(1+2+3+ldots+12) = 1,560 m$$

We see that $D_L+D_R = 3,000 m$.

Your error arises from assuming that the worker starts in the middle. It is more efficient to start at one end and carry rocks to the middle.

Assume that there are $n$ rocks on each side of the middle rock. If the worker starts at the end, he has to carry that rock a distance of $10n~text{m}$. For the worker to move the $k$th rock on that side to the middle, where $1 leq k leq n - 1$, the worker must travel $10k~text{m}$ to the rock and carry the rock $10k~text{m}$ back to the middle. On the opposite side, the worker must travel $10k~text{m}$ to the rock, then carry the rock $10k~text{m}$ back to the middle, where $1 leq k leq n$. Since the worker travels a total of $3000~text{m}$,
begin{align*}
10n~text{m} + sum_{k = 1}^{n - 1} 20k~text{m} + sum_{k = 1}^{n} 20k~text{m} & = 3000~text{m}\
n + 2sum_{k = 1}^{n - 1} k + 2sum_{k = 1}^{n} k & = 300\
n + 2sum_{k = 1}^{n - 1} k + 2sum_{k = 1}^{n - 1} k + 2n & = 300\
3n + 4sum_{k = 1}^{n - 1} k & = 300\
3n + 4 cdot frac{n(n - 1)}{2} & = 300\
3n + 2n(n - 1) & = 300\
2n^2 + n & = 300\
2n^2 + n - 300 & = 0\
2n^2 + 25n - 24n - 300 & = 0\
n(2n + 25) - 12(2n + 25) & = 0\
(n - 12)(2n + 25) & = 0
end{align*}
Hence, $n = 12$ or $n = -25/2$. Since we cannot have a negative number of rocks, there must be $12$ rocks on each side of the middle rock, so there are a total of $2 cdot 12 + 1 = 25$ rocks in the row.