Q.In a potato race $20$ potatoes are placed in a line at intervals of $4$ metres with the first potato $24$ metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Concept: Arithmetic progression — the distances form an AP because each potato is 4 m further than the previous one, and the contestant runs to a potato and back.

Step 1: The first potato is 24 m from the start. Distance to fetch it and return = $2 \times 24 = 48$ m.

Step 2: Each subsequent potato is 4 m farther. So the round-trip distances are: $48, 56, 64, \dots$ up to the 20th term. …

Why this works

The problem is a classic application of arithmetic progressions in a real-world context. Each potato is at a fixed distance from the start, and the contestant must run to that potato and back — so each trip is a round trip of twice the one-way distance. The one-way distances themselves form an arithmetic progression: the first potato is 24 m away, the next is 28 m away (24 + 4), then 32 m, and so on. The total distance is simply the sum of all these round-trip distances.

The key insight: you don't need to simulate 20 trips. Once you recognise the pattern, you can use the sum formula for an AP.

Step-by-step solution

1. Find the one-way distance to each potato.

The first potato is 24 m from the start. Each subsequent potato is 4 m further. So the one-way distances (in metres) are:

This is an arithmetic progression with first term $a = 24$ and common difference $d = 4$.

2. Find the distance to the 20th potato.

The $n$th term of an AP is $a + (n-1)d$. For $n = 20$:

So the farthest potato is 100 m from the start.

3. Compute the total one-way distance for all 20 potatoes.

The sum of the first $n$ terms of an AP is:

Here $n = 20$, first term $= 24$, last term $= 100$:

$S_{20} = \frac{20}{2} \times (24 + 100) = 10 \times 124 = 1240 \text{ m}$ …