Question -
Answer -
Given: totaltrees are 25 and equal distance between two adjacent trees are 5 meters
We need to find thetotal distance the gardener will cover.
As gardener is comingback to well after watering every tree:
Distance covered bygardener to water 1st tree and return back to the initialposition is 10m + 10m = 20m
Now, distance betweenadjacent trees is 5m.
Distance covered byhim to water 2nd tree and return back to the initial positionis 15m + 15m = 30m
Distance covered bythe gardener to water 3rd tree return back to the initialposition is 20m + 20m = 40m
Hence distance coveredby 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, n = 25
We need to find S25 whichwill be the total distance covered by gardener to water 25 trees.
So by using theformula,
Sn =n/2 [2a + (n – 1)d]
S25 =25/2 [2(20) + (25-1)10]
= 25/2 [40 + (24)10]
= 25/2 [40 + 240]
= 25/2 [280]
= 25 [140]
= 3500
∴ The total distancecovered by gardener to water trees all 25 trees is 3500m.