Question

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 3km.find the no. Of stones.

Answer 1

||✪✪ QUESTION ✪✪||

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 3km.find the no. Of stones. [ Excellent Question. ]

|| ★★ FORMULA USED ★★ ||

➪ An odd Number can be Written in the form of (2n +1) .

➪ Sum of n terms of AP is :- (n/2) [ 2a + (n-1)d ] , where a is first Term of AP, and d is common Difference ..

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image First .. ❁❁

From image we can see that, , Their were Total (2n+1) Stones, i assumed 1 middle Stone at Point B, n stones on the left of B and n stones on the Right of B.

______________________

Now, From image , we get , ,

☛ 20n² + 10n = 3000

☛ 20n² + 10n - 3000 = 0

Taking 10 common ,

☛ 10(2n² + n - 300) = 0

☛ 2n² + n - 300 = 0

Splitting The Middle Term now,

☛ 2n² - 24n + 25n - 300 = 0

☛ 2n(n - 12) + 25(n - 12) = 0

☛ (n - 12)(2n + 25) = 0

Putting both Equal to Zero now,

☛ n - 12 = 0. or, ☛ 2n + 25 = 0

☛ n = 12. ☛ n = (-25/2) .

As, Negative Value not Possible.

So, ☙ n = 12. ☙

So, Total Number of Stones = (2n+1) = (2*12 + 1) = 24 + 1 = 25 .

Hence, we can say that there are 25 Stone Total along side of the road.

Answer 2

So, after getting the A.P sequence shown in the attachment above . We'll find the sum of that A.P .

$4( \frac{n}{2} (20 + (n - 1)10) - 10n = 3000$

$2n(20 + 10n - 10) - 10n= 3000$

$2n(10 + 10n) - 10n = 3000$

$20n + 20 {n}^{2} - 10n = 3000$

$20n(1 + n) - 10n = 3000$

$2n(1 + n) - n = 300$

$2 {n}^{2} + 2n - n = 300$

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

$2 {n}^{2} + 25n - 24n - 300 = 0$

$n(2n + 25) - 12(2n - 25) = 0$

$(n - 12)(2n + 25) = 0$

$n = 12 \: or \: n = \frac{ - 25}{2}$

$n = - \frac{25}{2} is \: not \: possible \: \: so \: n = 12$

Total number of stones will = 2n+1

→ 2(12)+1 = 25 .