Solution:

It is given that total number of trees are 25 and the distance between two adjacent trees are 5 meters
To find the total distance the gardener will cover.
As given the gardener is coming back to well after watering every tree:

The distance covered by gardener to water $1^{st}$ tree and return back to the initial position is $10m + 10m = 20m$

So now, the distance between adjacent trees is 5m.

The distance covered by him to water $2^{nd}$ tree and return back to the initial position is $15m + 15m = 30m$

The distance covered by the gardener to water $3^{rd}$ tree return back to the initial position is $20m + 20m = 40m$

As a result, the distance covered by the gardener to water the trees are in the form of A.P

A.P. is $20, 30, 40 \dots \dots upto 25 terms$

Here, the first term, $a = 20$, the common difference, $d = 30 – 20 = 10$, $n = 25$

We now have to find $S_{25}$ which will be the total distance covered by gardener to water 25 trees.

Using the formula,
$\begin{array}{l}
S_{n}=n / 2[2 a+(n-1) d] \\
S_{25}=25 / 2[2(20)+(25-1) 10] \\
=25 / 2[40+(24) 10] \\
=25 / 2[40+240] \\
=25 / 2[280] \\
=25[140] \\
=3500
\end{array}$
As a result, the total distance covered by gardener to water trees all 25 trees is $3500 \mathrm{~m}$.