Question

Let there be odd number of stones placed one by one at an interval of $10\text{m}$ along a straight road. All the stones has to be assembled at the middle stone. A person start from one end and can only carry one stone at a time. If the distance covered by the person is $3\text{km}$ in this job, then the number of stones is

Answer 1

Let there be $2n+1$ stones. So, there will be $n$ stones each side of middle stone.
If the person started the job from middle,
The distance covered for first stone
$=2\times 10\text{m}$
Similarly, distance covered for second stone
$=2\times 20\text{m}$

So, the distance covered in bringing stones to centre from one side
$=2[10+20+30+\cdots +10n]$

Distance covered in bringing stones from both sides
$=4[10+20+30+\dots 10n]$

Since, the person started from one end, the total distance covered
$3000=4[10+2\times 10+3\times 10+...+n\times 10]-n\times 10\Rightarrow 40[1+2+3+...+n]-10n=3000\Rightarrow 4\times \frac{n(n+1)}{2}-n=300\Rightarrow 2n^2+n-300=0\Rightarrow (2n+25)(n-12)=0\therefore n=12(\because n\in N)$

Hence, number of stones is $2n+1=25$.