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.

Solution:

We have:

the distance travelled to bring the first potato, $a_{1}=2 \times 24=48 \mathrm{~m}$,

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

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

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, $a_{1}, a_{2}, a_{3}, \ldots$ 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.