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In a potato race 20 potatoes are placed in a line at intervals of $4 \mathrm{~m}$ with the first potato $24 \mathrm{~m}$ 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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Verified Answer
$$
\text { Given situation can be shown be following diagram }
$$

Distance travelled by the contestant to bring first potato $=24+24=48 m$
Distance travelled by the contestant to bring second potato $=2(24+4)=56 m$
Distance travelled by the contestant to bring third potato
$$
=2(24+4+4)=64 m
$$
Then, the series becomes: $48,56,64, \ldots$
Which is an AP.
Where, $a=48$,
$d=56-48=8$
and $n=20$
Required distance is equal to the sum of 20 terms of this series.
$$
\begin{aligned}
&\therefore S_{20}=\frac{20}{2}[2 \times 48+19 \times 8] \\
&=10[96+152] \\
&=10 \times 248=2480 \mathrm{~m}
\end{aligned}
$$
\text { Given situation can be shown be following diagram }
$$

Distance travelled by the contestant to bring first potato $=24+24=48 m$
Distance travelled by the contestant to bring second potato $=2(24+4)=56 m$
Distance travelled by the contestant to bring third potato
$$
=2(24+4+4)=64 m
$$
Then, the series becomes: $48,56,64, \ldots$
Which is an AP.
Where, $a=48$,
$d=56-48=8$
and $n=20$
Required distance is equal to the sum of 20 terms of this series.
$$
\begin{aligned}
&\therefore S_{20}=\frac{20}{2}[2 \times 48+19 \times 8] \\
&=10[96+152] \\
&=10 \times 248=2480 \mathrm{~m}
\end{aligned}
$$
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