Along a road lie an odd number of stones placed at intervals of 10 metres. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried 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.

Explanation

Let the number of stones =n
The distance between every stone= 10 m

So, here an AP is formed , a=20, d=20

= `n/2 [2a+(n−1)d]`

Distance between in both side,

⇒`4800=2( n/2 2a+(n−1)d)`

⇒`=n[2xx20+(n−1)20]`

∴ `n=15` (on one side)

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