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?
Answers
Hence the total distance to run is 2480 mtrs
Explanation :
The 1st potato is placed at 24 mtrs from the starting point
=> Distance to run to bring the 1st potato = 24 x 2 = 48
The 2nd potato is placed at (24+4) mtrs from the starting point
=> Distance to run to bring the 1st potato = (24+4) x 2 = 48 + 8
The 3rd potato is placed at (24+8) mtrs from the starting point
=> Distance to run to bring the 1st potato = (24+8) x 2 = 48+16
continuing this pattern we see the total distance to run
= 48 + (48+8) + (48+16) + (48+3x8).......................... + (48+ 19x8)
= 20 x 48 + ( 8 + 2x8 + 3x8 ..........+19 x 8)
= 960 + 8( 1+2+3 ........+19)
= 960 + 8(19 x 20/2) (as sum of arithmatic series = n*(n+1)/2)
= 960 + 8 x 190
= 2480
Hence the total distance to run is 2480 mtrs
Answer : 2480 meters
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Step by step solution :
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Given that : There are total 20 potatoes in a line at the interval of 4 meters.
First potato 24 metres from the start line. There area unit four meters within the intervals. A contestant is needed to bring the potatoes back to the root one at a time. thus for the primary potato he needs to travel 48 meters, for second fifty 56 meters
As we can get from here the series will be :
48, 56, 64, ......
The terms would be 20 so, n = 20
a = 48, d = 56-48 = 8
As we can see that the series represents the A.P and we know that : In A.P. sum of n terms
Sn = n/2[2a+(n-1)d]
On putting the value of following terms :
=> S20 = 20/2[2×48+(20-1)8]
=> S20 = 10[96+19×8]
=> S20 = 10[96+152]
=> S20 = 10×248
=> S20 = 2480
Thus, He would run 2480 meters in bringing back all the potatoes.