In a potato race $20$ potatoes are placed in a line at 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?

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Solution

For the first potato, contestant will cover

= $24 + 24 = 48 m$ .

$(∵$ Contestant is at starting place)

For the second potato, contestant will cover = $28 + 28 = 56 m$ .

For the third potato, contestant will cover

= $32 + 32 = 64 m$ .

and so on …

So, Sequence formed by distances covered for each potato is : $48, 56, 64$ , …

Here, $56 - 48 = 64 - 56 = 8$

As the difference of consecutive terms is constant the sequence is an arithmetic progression.

We have, first term $a = 48$ ,

common difference $d = 8$ .

Total distance covered by contestant for $20$ potatoes will be equal to sum of first $20$ terms of this AP.

$\Rightarrow$ Sum of n terms of an AP is given by

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\Rightarrow S_{20} = \frac{20}{2} [2 \times 48 + (20 - 1) d]$

$\Rightarrow S_{20} = 10 [96 + 152]$

$\Rightarrow S_{20} = 10 \times 248$

$\Rightarrow S_{20} = 2480 m$

$∴$ The total distance covered by a contestant to finish the race is equal to $2480 m$ .

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