NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3

Question 20

Idea: For each potato, the competitor runs from the bucket to the potato and back to the bucket. So, if a potato is at distance \(d\) metres, the distance run for that potato is \(2d\) metres. We add this for all 10 potatoes.

Step 1: Write the distances of the potatoes from the bucket

The first potato is 5 m away. Each next potato is 3 m further than the previous one.

So the distances form an AP:

\[5, 8, 11, 14, \ldots\]

Here, first term \(a = 5\), common difference \(d = 3\), and number of potatoes \(n = 10\).

Step 2: Find the sum of distances to all potatoes

Sum of first \(n\) terms of an AP:

\[S_n = \frac{n}{2}[2a + (n - 1)d]\]

For \(n = 10\), \(a = 5\), \(d = 3\):

\[S_{10} = \frac{10}{2}[2 \cdot 5 + 9 \cdot 3]\]

\[S_{10} = 5[10 + 27] = 5 \cdot 37 = 185\]

So, the sum of all one-way distances to the potatoes is 185 m.

Step 3: Account for going and coming back

For each potato, the competitor goes to the potato and comes back to the bucket. So the distance for each potato is doubled.

Total distance run:

\[\text{Total distance} = 2 \times S_{10} = 2 \times 185 = 370\ \text{m}\]

Conclusion

The competitor runs a total distance of \(370\text{ m}\).