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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.


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Solution

Let the number of stones be, 2n+1

From given we have and can calculate,

Sn=n2[2a+(n1)d]

=n2[2(10)+(n1)10]

=5n2+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[10(1)+10(2)+10(3)++10(n1)+10(n)]

Total distance

=2[10(1)+10(2)+10(3)++10(n1)]+10(n)+2[10(1)+10(2)+10(3)++10(n1)+10(n)]

=4[10(1)+10(2)+10(3)++10(n1)]+30n]

=4[10(1)+10(2)+10(3)++10(n1)+10(n)]10n

But the total distance covered is 3km or 3000 m

Hence 4[10(1)+10(2)+10(3)++10(n1)+10(n)]10n=3000

4[5n2+5n]10n=3000

20n2+20n10n=3000

20n2+10n3000=0

(2n+25)(n12)=0

n=12

Therefore the number of stones =2(12)+1=25


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