Question

Along a road lie an odd number of stones and distance between consecutive stones is $10m$. A person can carry only one stone at a time and his job is to assemble all the stones around the middle stone. If he starts his job from one of the end stones and is carrying all the stones, he covers a distance of $3km$. Find the number of stones.

Answer 1

Hint- In order to deal with this question first we will assume number of stones as a variable and then we will evaluate distance covered in collecting all the stones on the left side after reaching at middle stone we will repeat same process which we will do for left side further by getting total distance an AP will formed and by using the sum of n terms of AP we will get the number of stones.

Complete step-by-step answer:
Let the number of stones be $(2n+1)$ since there are an odd number of stones.

Given these stones are to be kept around the middle stone.

That is there will be $n$ stones kept on the 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 $1$st to $n$ stones and drops it around the middle stone.

The distance covered in collecting all the stones on the left side $2\left[10(1)+10(2)+10(3)+..........10(n-1)\right]+10(n)$

$\because$ the man was initially at the extreme left then $n^{th}$ 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 on the 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
$=2\left[10(1)+10(2)+.........10(n-1)+10(n)\right]$

Therefore, Total distance = The distance covered in collecting all the stones on the left side $+$ the distance covered on the right side

$$\begin{gathered} =2\left[10(1)+10(2)+10(3)+..........10(n-1)\right]+10(n)+2\left[10(1)+10(2)+.........10(n-1)+10(n)\right] \\ =4[10(1)+10(2)+10(3)+.........10(n-1)+30(n)] \\ =4[10(1)+10(2)+10(3)+.......10(n-1)+10(n)]-10(n) \end{gathered}$$


But the total distance covered is $3km\text{ or }3000m$

Hence, $=4[10(1)+10(2)+10(3)+.......10(n-1)+10(n)]-10(n)=3000\to (1)$

$$\begin{gathered} =10(1)+10(2)+10(3)+.....10(n-1)+10(n) \\ =10+20+30+.....10(n-1)+10(n) \end{gathered}$$

Here $$a=10,d=10$$

Recall the sum of $$n$$ terms of AP, $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$
​
 Substitute the values in above formula we have

$$\begin{gathered} S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d] \\ ={\displaystyle \frac{n}{2}}[2(10)+(n-1)10] \\ ={\displaystyle \frac{n}{2}}[20+10n-10] \\ ={\displaystyle \frac{n}{2}}[10+10n] \\ =5n^2+5 \end{gathered}$$

Equation (1) becomes as:

$$\begin{gathered} 4[5n^2+5n]-10n=3000 \\ =20n^2+20n-10n=3000 \\ =20n^2+20n-10n-3000=0 \end{gathered}$$

Further by solving above quadratic equation we have

$$\begin{gathered} =20n^2+10n-3000=0 \\ =10(2n^2+2n-300)=0 \\ =(2n^2+2n-300)=0 \\ =n(2n+25)-12(2n+25)=0 \\ =(2n+25)(n-12)=0 \\ =(2n+25)=0\text{ or }(n-12)=0 \\ \therefore n=12\text{ or n = }{\displaystyle \frac{-25}{2}} \end{gathered}$$

But $n$cannot be negative or fraction
So, $n=12$

Therefore, number of stones $$=2n+1=2\left(12\right)+1=25$$

Note- The sum of $$n^{th}$$term in A.P. is given by the formula $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$ where $$a$$is the first term of A.P., $$d$$is the common difference of the two consecutive terms which remains same and $$n$$is the total terms present in A.P.