provide answer the following question with procedure
Answers
Answer. 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 m Hence 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 = 10 Recall 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 = 12 Therefore, number of stones = 2n + 1 = 2(12) + 1 = 25