Question

Along a road side an odd number of stones placed at intervals of 10 meters. These stones have to be assembled around the middle stone. A person can carry one stone at a time. A man carries the job with one of the end stones by carrying them in succession. In carrying all the stones he covered a distance of 3 km. Find the number of stones.

Answer 1

Let the number of stones be, $2n+1$

From given we have and can calculate,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}\left[2\right(10)+(n-1\left)10\right]$

$=5n^2+5n$

given, $3km$

The difference is that here he will cover distance both the way in collecting all the stones.

Hence the distance covered on the right side $=2\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]$

Total distance

$=2\left[10\right(1)+10(2)+10(3)+\dots +10(n-1\left)\right]+10\left(n\right)+2\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]$

$=4\left[10\right(1)+10(2)+10(3)+\dots +10(n-1\left)\right]+30n]$

$=4\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]–10n$

But the total distance covered is 3km or 3000 m

Hence $4\left[10\right(1)+10(2)+10(3)+\dots +10(n-1)+10(n\left)\right]–10n=3000$

$4[5n^2+5n]-10n=3000$

$20n^2+20n-10n=3000$

$20n^2+10n-3000=0$

$(2n+25)(n-12)=0$

$\therefore n=12$

Therefore the number of stones $=2\left(12\right)+1=25$