Question

in a Potato Race 20 potatoes are placed in a line of interval of 4m with the first potato 24m 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

Answer 1

Hey......

Given, total number of potatoes = 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.

Answer 2

To pick the first potato, the contestant has to run 24m to reach the potato and 24m to run back to the starting place.
so, total distance covered by the contestant to pick the first potato = 2 × 24 = 48m

To pick the 2nd potato, the contestant has to run (24 + 4)m to reach potato and (24 + 4)m to run back to the starting place.
so, total distance covered by the contestant to pick the 2nd potato = 2(24 + 4) = 56m

To pick the 3rd potato, the contestant has to run (24 + 4 + 4)m to reach potato and (24 + 4 + 4)m to run back to the starting place.
so, total distance covered by the contestant to pick the 3rd potato = 2(24 + 4 + 4) = 64m

this will continue and we will get a sequence of distance as , 48, 56, 64, 72, .... upto 20 terms

so, total distance covered by the contestant to pick all the 20 potatoes = 48 + 56 + 64 + 72 .....+ upto 20 terms

here you observed, sequence is in AP
where first term, a = 48
and common difference , d = 8
and number of terms, n = 20

use sum of n terms in AP
e.g., $S_n=\frac{n}{2}[2a+(n-1)d]$

= 20/2 [ 2 × 48 + (20 - 1) × 8]

= 10[ 96 + 19 × 8 ]

= 10[ 96 + 152]

= 2480