Question

A long a road lie an odd number of stone placed at intervals of $10m$. These stones have to be assembled around the middle stone. A man carried the job with one of the end stone by carrying them in succession. In carrying all the stone he covered a distance of $3km$. Then how many stones were there?

Answer 1

The distance covered on the right side

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

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

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

$\Rightarrow$ Total distance covered $3000$

$a=10$, $d=10$

Sum $\frac{n}{2}\left[2\right(10)+(n-1\left)10\right]=3000$

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

$2n^2+n-300=0$

$n=12$, $n=-\frac{25}{2}$

Total no. of stones$=2n+1$

$=2\left(12\right)+1=25$.