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