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`        Along a road lie an odd number of stones placed at intervals of 20metres. these stones have to be assembled around the middlw stone. a person carry only 1 stone at a time. A man carried his job with one of the end stones by carrying them in succession. In carrrying all the stones he covered a distance of 3km. Find the number of stones? plss can someone provide a solution WITH PROPR EXPLAINATION , i just nver understand it.`
2 years ago

```							we may use the sum of ap
because there is an arrangement of stone in the form ap
First count the no of points at interval of 20mts in 3km route
then we do
3*1000m/20=150 points
then the no of stones on each point may be like this
1,3,5,7,9...
find out last no of stones by formula of ap
Tn=a+(n-1)d
d=3-1=2
a=1
n=150
Tn=1+(150-1)2
=1+149*2
=299
Then apply formula of Sn=n/2(2a+(n-1)d)
=150/2(2*1+(150-1)2)
=75(2+298)
=75*300
=22500 total no of stones. Answer
```
one year ago
```							Let the number of stones be (2n + 1) since there are odd number of stones.Given these stones are to be kept around the middle stone.That is there will n stones kept on right side of middle stone and n stones kept on the left side of the middle stone.It is also given that there are n intervals each 10m on both the sides.If the man collects the stones from 1st to n stones and drops it around the middle stone.The distance covered in collecting all the stones on the left side = 2[10(1) + 10(2) + 10(3) +… + 10(n-1)] + 10(n)  [Because the man was initially at the extreme left then nth stone he will cover one way distance whereas other stones he has cover two way distance]Now the person is at the middle stone. Now repeats the same process as done in collecting the stones in left side. 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(n-1) + 10(n)]Total distance = {2[10(1) + 10(2) + 10(3) +… + 10(n-1)] + 10(n)} + 2[10(1) + 10(2) + 10(3) +… + 10(n-1) + 10(n)]= 4[10(1) + 10(2) + 10(3) +… + 10(n-1)] + 30n]= 4[10(1) + 10(2) + 10(3) +… + 10(n-1) + 10(n)] – 10n But the total distance covered is 3km or 3000 mHence 4[10(1) + 10(2) + 10(3) +… + 10(n-1) + 10(n)] – 10n = 3000  →  (1)But 10(1) + 10(2) + 10(3) +… + 10(n-1) + 10(n) forms an AP.Here a = 10, d = 10Recall the sum of n terms of AP, Equation (1) becomes,4[5n2 + 5n] – 10n = 3000⇒ 20n2 + 20n – 10n = 3000⇒ 20n2 + 10n – 3000 = 0⇒ 2n2 + n – 300 = 0⇒ 2n2 + 25n – 24n – 300 = 0⇒ n(2n + 25) – 12(2n + 25) = 0⇒ (2n + 25)(n – 12) = 0⇒ (2n + 25) = 0 or (n – 12) = 0∴ n = 12 or n = (– 25/2)But n cannot be negative or fraction⇒ n = 12Therefore, number of stones = 2n + 1 = 2(12) + 1 = 25
```
one year ago
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