Answer :

**Given:** total trees are 25 and equal distance between two adjacent trees are 5 meters

**To find:** the total distance the gardener will cover

As gardener is coming back to well after watering every tree

Distance covered by him to water 1^{st} tree and return to the initial position is 10m + 10m = 20m

Now, distance between adjacent trees is 5m

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

Distance covered by the gardener to water 3^{rd} tree return to the initial postion is 20m + 20m = 40m

Hence distance covered by the gardener to water the trees are in A.P

A.P. is 20, 30, 40 ………upto 25 terms

Here first term,a = 20,common difference, d = 30 – 20 = 10

And n = 25

We need to find S_{25} which will be the total distance covered by gardener to water 25 trees

**Formula used:**

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

⇒ S_{25} = 25(140)

⇒ S_{25} = 3500

**Hence, total distance covered by gardener to water trees is 3500 m**

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