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Along a road lie an odd number of stones and distance between consecutive stones is $10{\text{m}}$. 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.

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: Here, we will use the given information to find an arithmetic sequence. Then we will use the sum of $n$ terms in an AP to solve it further and find the value of $n$. Then we will substitute this value in the formula of an odd number of stones to find the required value of the number of stones. An Arithmetic Progression or A.P. is a sequence in which the difference between two consecutive terms is the same.

Formula Used:
The sum of $n$terms in an AP is given as ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
Where, the first term is $a$ and the common difference is $d$.

Complete step-by-step answer:
According to the question, there are an odd number of stones lying on the road.
Hence, let the number of stones be $\left( {2n + 1} \right)$
Now, these stones are to be kept around the middle stone.
Hence, there will be $n$ stones on the right side of middle stone and $n$ stones on the left side of the middle stone.
Also, the distance between consecutive stones is $10{\text{m}}$.
And, there are $n$ intervals each $10{\text{m}}$ on both the sides.
If the man collects the stones from first to $n$ stones and drops it around the middle stone.
The distance covered in collecting all the stones on the left side of the middle stone$ = 2\left[ {10\left( 1 \right) + 10\left( 2 \right) + 10\left( 3 \right) + ... + 10\left( {n - 1} \right)} \right] + 10\left( n \right)$
[We have multiplied this by 2 because the man was initially at the extreme left and he can collect only one stone at a time. Thus, he has to cover two way distances for each stone except the ${n^{th}}$ stone]
Now, the person is at the middle stone. He 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\left( 1 \right) + 10\left( 2 \right) + 10\left( 3 \right) + ... + 10\left( {n - 1} \right) + 10\left( n \right)} \right]$
Therefore,
Total distance $ = 2\left[ {10\left( 1 \right) + 10\left( 2 \right)+... + 10\left( {n - 1} \right)} \right] + 10\left( n \right) + 2\left[ {10\left( 1 \right) + 10\left( 2 \right) ... + 10\left( {n - 1} \right) + 10\left( n \right)} \right]$
$ \Rightarrow $ Total distance $ = 4\left[ {10\left( 1 \right) + 10\left( 2 \right) ... + 10\left( {n - 1} \right)} \right] + 10\left( n \right) + 20\left( n \right)$
$ \Rightarrow $ Total distance $ = 4\left[ {10\left( 1 \right) + 10\left( 2 \right) + ... + 10\left( {n - 1} \right)} \right] + 30\left( n \right)$
Now, this can also be written as:
$ \Rightarrow $ Total distance $ = 4\left[ {10\left( 1 \right) + 10\left( 2 \right) + 10\left( 3 \right) + ... + 10\left( {n - 1} \right) + 10\left( n \right)} \right] - 10\left( n \right)$
But according to the question, the total distance covered is $3{\text{km}}$ or $3 \times 1000 = 3000{\text{m}}$
Hence, we get,
$4\left[ {10\left( 1 \right) + 10\left( 2 \right) + 10\left( 3 \right) + ... + 10\left( {n - 1} \right) + 10\left( n \right)} \right] - 10\left( n \right) = 3000$……………….$\left( 1 \right)$
But, clearly we can observe that the above sequence is an arithmetic progression.
Hence, here first term, $a = 10$
Common difference, $d = 10$
Also, the sum of $n$ terms in an AP is given as ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Hence, substituting the above values, we get,
${S_n} = \dfrac{n}{2}\left[ {2\left( {10} \right) + \left( {n - 1} \right)\left( {10} \right)} \right] = \dfrac{n}{2}\left[ {20 + 10\left( {n - 1} \right)} \right]$
Taking 10 common and dividing it by 2 which is in the denominator, we get,
${S_n} = \dfrac{n}{2} \times 10\left[ {2 + \left( {n - 1} \right)} \right] = 5n\left[ {2 + n - 1} \right] = 5n\left[ {n + 1} \right]$
Thus, we get,
${S_n} = 5{n^2} + 5n$
Thus, equation $\left( 1 \right)$ becomes,
$4\left( {5{n^2} + 5n} \right) - 10n = 3000$
Multiplying the bracket by 4, we get
$ \Rightarrow 20{n^2} + 20n - 10n = 3000$
$ \Rightarrow 20{n^2} + 10n = 3000$
Dividing both sides by 10, we get,
$ \Rightarrow 2{n^2} + n = 300$
Subtracting 300 from both the sides, we get
$ \Rightarrow 2{n^2} + n - 300 = 0$
Doing middle term split, we get
$ \Rightarrow 2{n^2} + 25n - 24n - 300 = 0$
$ \Rightarrow 2n\left( {n - 12} \right) + 25\left( {n - 12} \right) = 0$
Factoring out common terms, we get
$ \Rightarrow \left( {2n + 25} \right)\left( {n - 12} \right) = 0$
Now using zero product property, we get
$2n + 25 = 0$
$ \Rightarrow n = \dfrac{{ - 25}}{2}$
Also,
$n - 12 = 0$
$ \Rightarrow n = 12$
But $n$ cannot be negative or in fraction.
Thus, $n = 12$

Therefore, number of stones $ = 2n + 1 = 2\left( {12} \right) + 1 = 24 + 1 = 25$
Hence, the required number of stones is 25.
Thus, this is the required answer.


Note:
The difference in AP series is known as the common difference and we find it by subtracting each term by its preceding term respectively. Arithmetic progressions are also used in real life such as adding the same amount as our pocket money in our money bank. Since, we add the same amount each time, that amount or that pocket money will become our common difference in this case. Similarly, we hire a taxi, we are charged an initial rate and then rate per kilometer. That rate per kilometer becomes our common difference and each addition gives us an A.P. Hence, this is used in our day to day life.