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

We have,

The distance travelled to bring the first potato, $a_{2} = 2 \times 24 = 48 m$

The distance travelled to bring the second potato, $a_{2} = 2 \times (24 + 4) = 56 m$

The distance travelled to bring the third potato, $a_{3} = 2 \times (24 + 4 + 4) = 64 m$

As, $a_{2} - a_{1} = 56 - 48 = 8$ and $a_{3} - a_{2} = 64 - 56 = 8$

i.e., $a_{2} - a_{1} = a_{3} - a_{2}$

So, $a_{1}, a_{2}, a_{3}, \dots \dots$ are in A.P.

Also, a = 48, d = 8, n = 20

Now,

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

$= 10 [2 \times 48 + 19 \times 8]$

$= 10 \times (96 + 152)$

$= 10 \times 248 = 2480$

So, he would have to run 2480 m to bring back all the potatoes.

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In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

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