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(A) The distances of the potatoes from the bucket are $5\, m, 8\, m, 11\, m, 14\, m, \dots$ This forms an Arithmetic Progression $(A.P.)$ where the fi

Vedclass · Accepted Answer

$370$

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(A) The distances of the potatoes from the bucket are $5\, m, 8\, m, 11\, m, 14\, m, \dots$ This forms an Arithmetic Progression $(A.P.)$ where the first term $a = 5$ and the common difference $d = 3$. For each potato,the competitor runs to the potato and back to the bucket,so the distance covered for each potato is twice its distance from the bucket. The distances to be run for each of the $10$ potatoes are $2 	imes 5, 2 	imes 8, 2 	imes 11, 2 	imes 14, \dots$ This is a new $A.P.$ with first term $a' = 10$ and common difference $d' = 6$. The total distance is the sum of the first $10$ terms of this $A.P.$ Using the sum formula $S_n = \frac{n}{2}[2a' + (n-1)d']$: $S_{10} = \frac{10}{2}[2(10) + (10-1)6]$ $S_{10} = 5[20 + 9 	imes 6]$ $S_{10} = 5[20 + 54]$ $S_{10} = 5[74] = 370$. Thus,the total distance the competitor has to run is $370\, m$.

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