Along a rode lies an odd …

Lets suppose there are (2n + 1) stones. Clearly, one stone lies in the middle and n stones on each side of it in a row. Let P be the mid-stone and let A and B be the end stones on the left and right of P respectively.

Clearly, there are n intervals each of length 10 metres on both the sides of P. Now, suppose the man starts from A. He picks up the end stone on the left of mid-stone and goes to the mid-stone, drops it and goes to (n - 1)th stone on left, picks it up, goes to the mid-stone and drops it. This process is repeated till he collects all stones on the left of the mid-stone at the mid-stone. So, distance covered in collecting stones on the left of the mid-stones is
10 {tex} \times{/tex} n + 2 [10 {tex} \times{/tex} (n - 1) + 10  {tex} \times{/tex} (n - 2) + ... + 10 {tex} \times{/tex} 2 + 10 {tex} \times{/tex} 1].After collecting all stones on left of the mid-stone the man goes to the stone B on the right side of the mid-stone, picks it up, goes to the mid-stone and drops it. Then, he goes to (n - 1)^th stone on the right and the process is repeated till he collects all stones at the mid-stone.So that distance covered in collecting the stones on the right side of the mid-stone is equal to 2 [10 {tex} \times{/tex} n + 10 {tex} \times{/tex} (n - 1) + 10 {tex} \times{/tex} (n - 2)+ ... +10 {tex} \times{/tex}  2 + 10 {tex} \times{/tex} 1].
Therefore,total distance covered
= 10 {tex} \times{/tex} n + 2 [10 {tex} \times{/tex} (n - 1) + 10 {tex} \times{/tex} (n - 2) + ... + 10 {tex} \times{/tex} 2 + 10 {tex} \times{/tex} 1]+ 2 [10 {tex} \times{/tex}  n + 10 {tex} \times{/tex}  (n - 1) + ... + 10 {tex} \times{/tex}  2 + 10 {tex} \times{/tex} 1]= 4 [10 {tex} \times{/tex} n + 10 {tex} \times{/tex} (n - 1) + ... +10 {tex} \times{/tex} 2 + 10 {tex} \times{/tex} 1] -10 {tex} \times{/tex} n
{tex} = 40 \{ 1 + 2 + 3 + \ldots + n \} - 10 n = 40 \left\{ \frac { n } { 2 } ( 1 + n ) \right\} - 10 n = 20 n ( n + 1 ) - 10 n = 20 n ^ { 2 } + 10 n{/tex}But, the total distance  that a man covered in collecting stones  is 3 km.i.e; 3000 m.
Therefore, 20 n² + 10n = 3000 
{tex} \Rightarrow{/tex} 2n² + n - 300 = 0.{tex}\implies{2n^2+25n+24n-300=0}{/tex}
{tex} \Rightarrow{/tex} (n - 12) (2n + 25) = 0 [Therefore, 2n + 25 {tex} \neq{/tex} 0]{tex}\implies {n-12=0}{/tex}
{tex} \Rightarrow{/tex} n = 12.Hence, the number of stones is equal to 12.Which is the required answer.