Distances moved in bringing back the 1st, 2st,3rd,…..,20th potato to the starting point are ` ( 2 xx 24)m, ( 2xx 32)m`,…., and so on.
Total distance moved = { 48 + 56+64 +72 + ….to 20 term }
` = 20/2 xx [ 2 xx 48 + ( 20-1) xx 8]` m
[ a = 48 and n = 20]
` = 10 xx ( 96 + 152) m = 2480 m `
Hence , total distance covered = 2480 m