Question

In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3m apart in a straight line. There are n potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in the bucket and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Answer 1

Hint: In this question, the distance of potatoes and bucket along with way of picking and dropping potatoes is given. We have to calculate the total distance the competitor has to run to finish the race after picking n number of potatoes. For this, we will deduct distance from bucket to the first potato to the bucket and similarly distance from bucket to second potato to the bucket and so on. We will find that distance covered for first, second, third . . . . . . . ${{n}^{th}}$ potatoes forms an arithmetic progression, and hence we will use properties of AP to find the total distance covered. Total distance will be the sum of all terms (sum of all distance) and hence will be given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ where, a is the first term, d is the common difference and n is the number of terms.

Complete step-by-step solution:
Here, we are given the distance between the first potato and bucket as 5m. The competitor ran towards the potato, pick it, and came back to the bucket to drop it. So, the distance covered for one potato will be (5+5)m = 10m.
Now, for the second potato, the competitor will run towards the second potato placed at 3m from the first potato. So, the distance from the bucket to the second potato will be 5+3 = 8m. Distance for picking and dropping the second potato will become 8+8 = 16m.
Similarly, for third, the distance between the bucket and the third potato will be 8+3 = 11m, and the distance covered by the competitor to pick and drop potato will be 11+11 = 22m.
Hence, distance can be listed as 10, 16, 22 . . . . . . . .
As we can see, this series forms an arithmetic progression with the first term as 10 and the common difference as 16-10 = 22-16 = 6.
Therefore, a = 10 and d = 6.
Sum of n terms of an AP is given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
Here, a = 10 and d = 6. Competitor ran for n potatoes, therefore, n will be n only. Hence, total distance covered will be given as
\[\begin{align}
  & {{S}_{n}}=\dfrac{n}{2}\left( 2\times 10+\left( n-1 \right)6 \right) \\
 & = \dfrac{n}{2}\left( 20+6n-6 \right) \\
 & = \dfrac{n}{2}\left( 14+6n \right) \\
 &= n\left( 7+3n \right) \\
 & = 7n+3{{n}^{2}} \\
\end{align}\]
Hence, required distance covered is $3{{n}^{2}}+7n$.

Note: Students can make the mistake of considering the distance of one side only. Here, take care that the distance between the first potato and the bucket is different than the distance between consecutive potatoes. We have found distance for picking and dropping n potatoes which means we can put any value of potatoes in the answer to find the distance. For example, if there were 10 potatoes, distance covered would be $\Rightarrow 3{{\left( 10 \right)}^{2}}+7\left( 10 \right)=300+70=370m$.