**Question:**

**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?**

**Solution:**

Given at start he has to run 24m to get the first potato then 28 m as the next potato is 4m away from first and so on

Hence the sequence of his running will be 24, 28, 32 …

There are 20 terms in sequence as there are 20 potatoes

Hence only to get potatoes from starting point he has to run

24 + 28 + 32 + … up to 20 terms

This is only from starting point to potato but he has to get the potato back to starting point hence the total distance will be twice that is

Total distance ran = 2 × (24 + 28 + 32…) … 1

Let us find the sum using he formula to find sum of n terms of AP

That is Sn = n/2 (2a + (n – 1) d)

There are 20 terms hence n = 20

⇒ S20 = (20/2) (2 (24) + (20 – 1) 4)

On simplification we get

⇒ S20 = 10(48 + 19(4))

⇒ S20 = 10(48 + 76)

On computing we get

⇒ S20 = 10 × 124

⇒ S20 = 1240 m

Using equation 1

⇒ Total distance ran = 2 × 1240

⇒ Total distance ran = 2480 m

Hence total distance he has to run is 2480 m