In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters 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?

• sorry for the prev wrong ans..it will be..
  48+56+64+........upto 20th term
  so,
  s=n/2[2a+(n-1)d]=2480
• 9 years agoHelpfull: Yes(20) No(4)
• Explanation:
  Given, total number of potatos = 20.
  First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
  48,56,64...........20 terms.
  a = 48, d= 8, n = 20.
  Sum of n terms in A.P = Sn=n2[2a+(n−1)d]
  S20=202[2×48+(20−1)8]
  S20=202[96+152]
  S20=10×248=2480
  ∴ 2480 meters he run in bringing back all the potatoes.
• 9 years agoHelpfull: Yes(16) No(1)
• For every potato to bring it to iys starting point contestant has to run twice of the same distance and total distance as
  24*2+28*2+......100*2=2*4[(1+2+3+4+5+6+7+8+...25)-(1+2+3+4+5)]
  =8*[(25*(25+1)/2)-(5*(5+1)/2)]
  =8*[(25*13)-(5*3)]
  =8*310
  =2480 meters
  
• 9 years agoHelpfull: Yes(6) No(2)
• Given, total number of potatos = 20.
  First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
  48,56,64...........20 terms.
  a = 48, d= 8, n = 20.
  Sum of n terms in A.P = Sn=n2[2a+(n−1)d]
  S20=202[2×48+(20−1)8]
  S20=202[96+152]
  S20=10×248=2480
  ∴ 2480 meters he run in bringing back all the potatoes.
• 9 years agoHelpfull: Yes(4) No(0)
• Ans:- 2090. Asto bring back 1st potato he required 24+24 i.e. 48m, for 2nd 56m, 3rd 64m -------------- 20th 200m. Hence addition of 48, 56,64,72,----------192,200 is 2090.
• 9 years agoHelpfull: Yes(1) No(3)
• Why Do every one see something with so much difficulty ?
  
  1st potato =24 meters
  then all the potatoes are 4 meters appart = 19x4= 76meters
  therefore to collect all the potatoes, 24+76= 100 meters
  and to come back, = 100 more meeters
  
  thus total = 100 + 100 = 200 meters!
• 9 years agoHelpfull: Yes(1) No(9)
• 240..cz..48+56+64+72=240
  
• 9 years agoHelpfull: Yes(0) No(4)
• 24*2=48, 48*20=960, 8+16+24+...19terms = 8*(1+2+3+..+19)= 8*19*20/2=1520, 960+1520=2480 meters
• 9 years agoHelpfull: Yes(0) No(1)
• 24+(4*20)=104
  2*104=208
• 9 years agoHelpfull: Yes(0) No(4)
• @ ankit kumar verma.. could you please explain how did you get 2*4[(1+2+3+4+5+6+7+8+...25)-(1+2+3+4+5)] this term? I am little confused!!! :(
• 9 years agoHelpfull: Yes(0) No(1)
• we first apply the fourmula to find the last term ie. L=a+(n-1)d
  then sum of series we apply ie. sum=n/2(a+L)
  ans is 2480..
• 8 years agoHelpfull: Yes(0) No(3)
• please anyone can explain clearly
• 8 years agoHelpfull: Yes(0) No(0)
• Given, total number of potatos = 20.
  First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
  48,56,64...........20 terms.
  a = 48, d= 8, n = 20.
  Sum of n terms in A.P =
  Sn=n/2*[2a+(n−1)d]
  S20=20/2*[2×48+(20−1)8]
  S20=20/2*[96+152]
  S20=10×248=2480
  ∴ 2480 meters he run in bringing back all the potatoes.
• 5 years agoHelpfull: Yes(0) No(1)
• I think the answer is unsolved here, coz while running for second potato, he will go diagonally, we need to find the diagonal distance to reach to second potato by using phytogerous theorem.
• 5 years agoHelpfull: Yes(0) No(1)