In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6).Fig. 5.6A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
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To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is \(2 × 5 + 2 × (5 + 3)\).
The distances between successive potatoes are as follows: \(5, 8, 11, 14….\). This sequence is an arithmetic progression (A.P.) with first term \(a = 5\) and common difference \(d = 8 − 5 = 3\). The sum of the first \(n\) terms of an A.P. is given by \(S_n = \frac n2[2a + (n-1)d]\).
For the first 10 potatoes, the sum of distances is calculated as: \(S_{10} = \frac {10}{2}[2(5) + (10-1)3] = 5[10 + 9 × 3] = 5(10 + 27) = 5(37) = 185\). Since the competitor must return to the bucket after collecting each potato, the total distance run is twice this sum.
Therefore, the total distance the competitor runs is \(2 × 185 = 370 \ m\). Alternatively, The distances of the potatoes from the bucket are \(5, 8, 11, 14, ….\). The distance run by the competitor to collect each potato is double its distance from the bucket. Thus, the distances to be run form a new sequence: \(10, 16, 22, 28, 34,……\). This is an A.P. with first term \(a = 10\) and common difference \(d = 16 − 10 = 6\). We need to find \(S_{10}\).
