This is the basic and conceptual question from Chapter name- Arithmetic Progression

Chapter number- 9

Exercise 9.6

In this question we have been given that there are 25 trees at equal distances of 5 meter in a line with a well,

The distance of the well from the nearest tree being 10 meters.

A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next.

Now we are asked to find the total distance the gardener will cover in order to water all the trees.

This is very important question.

CBSE DHANPAT RAI publications

Class:- 10th

Solutions of CBSE Mathematics

Question 66

Given number of trees(n) = 25.

Distance covered by the gardener during watering of first tree = 2(10) = 20 meters

Distance covered by the gardener during watering of second tree = 2(10 + 5) = 30 meters

Distance covered by the gardener during watering of third tree = 2(10 + 5 + 5) = 40 meters

So, first term(a) = 20 and common difference(d) = 30 â€“ 20 = 10 meters.

Total distance covered to water 25 trees = Sum of 25 terms of A.P. = S

_{25}By using the formula of the sum of n terms of an A.P.

S

_{n}Â = n[2a +(n â€“ 1)d] / 2we get

= 25[2(20) + (25 â€“ 1)10] / 2

= 25[40 + 240] / 2

= 25[140]

= 3500 meters

Hence, the total distance the gardener will cover in order to water all the trees is 3500 meters.