# Ex.5.3 Q20 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

In a potato race, a bucket is placed at the starting point, which is \(5 \,\rm{m}\) from the first potato and other potatoes are placed \(3 \,\rm{m}\) apart in a straight line. There are ten 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, 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?

[**Hint:** to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is \(2 × 5 + 2 × (5 + 3)\)]

## Text Solution

**What is Known?**

Distance of \(10\) potatoes from the bucket, first potato is \(5 \,\rm{m}\) away from bucket and other potatoes are placed \(3\,\rm{m}\) apart in a straight line.

**What is Unknown?**

Total distance the competitor has to run.

**Reasoning:**

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

The distances of potatoes are as follows.

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

It can be observed that these distances are in A.P.

- First term, \(a = 5\)
- Common difference, \(d = 8 - 5 = 3\)
- Number of terms, \(n = 10\)

We know that the sum of \(n\) terms,

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{10}} &= \frac{{10}}{2}\left[ {2 \times 5 + (10 - 1) \times 3} \right]\\{S_{10}} &= 5\left[ {10 + 27} \right]\\{S_{10}} &= 5 \times 37\\{S_{10}} &= 185\end{align}\]

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run

\( = 2 \times 185\, \rm{m }= 370\,{\rm{m }}\)

**Alternatively, **

The distances of potatoes from the bucket are \(5, 8, 11, 14...\)

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, distances to be run are

First potato, \(a = 2 \times 5\,\rm{m} = 10\rm{m}\)

Second potato,

\[{a_2} = 2 \times \left( {5 + 3} \right)\,\rm{m} = 16\,\rm{m}\]

Third potato,

\[{a_3} = 2 \times \left( {5 + 2 \times 3} \right)\,\rm{m} = 22\,\rm{m}\]

Number of potatoes, \(n = 10\)

\(10,\,16,\,22,\,28,\,34,\ldots \ldots \ldots\)

\[\begin{align}a &= 10\\d& = 16 - 10 = 6\\{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{10}} &= \frac{{10}}{2}\left[ {2 \times 10 + (10 - 1)6} \right]\\&= 5\left[ {20 + 54} \right]\\&= 5 \times 74\\&= 370\end{align}\]

Therefore, the competitor will run a total distance of \(370 \,\rm{m}.\)