English

There Are 25 Trees at Equal Distances of 5 Metres in a Line with a Well, the Distance of the Well from the Nearest Tree Being 10 Metres. - Mathematics

Advertisements
Advertisements

Question

There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. 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. Find the total distance the gardener will cover in order to water all the trees.

Sum
Advertisements

Solution

In the given problem, there are 25 trees in a line with a well such that the distance between two trees is 5 meters and the distance between the well and the first tree is 10 meters.

So, the total distance covered to water first tree   = 10 meters

Then he goes back to the well to get water.

So,

The total distance covered to water second tree = 25 meters

The total distance covered to water third tree = 35 meters

The total distance covered to water fourth tree = 45 meters

So, from second tree onwards, the distance covered by the gardener forms an A.P. with the first term as 25 and common difference as 10.

So, the total distance covered for 24 trees can be calculated by using the formula for the sum of n terms of an A.P,

`S_n = n/2 [2a + (n-1)d]`

We get,

`S_n = 24/2 [2(25) + (24 - 1)(10)]`

      = 12 [ 50 +(23) (10)]

      = 12 (50 + 230 ) 

      = 12 (280)

      = 3360

So, while watering the 24 trees he covered 3360 meters. Also, to water the first tree he covers 10 meters. So the distance covered while watering 25 trees is 3370 meters.

Now, the distance between the last tree and the well 

= 10 + 24 (5) 

= 10 + 120 

= 130

So, to get back to the well he covers an additional 130 m. Therefore, the total distance covered by the gardener 

= 3370 + 130 

= 3500

Therefore, the total distance covered by the gardener is 3500 m .

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Arithmetic Progression - Exercise 5.6 [Page 54]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 66 | Page 54

RELATED QUESTIONS

Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.


 In an AP Given a12 = 37, d = 3, find a and S12.


How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?


The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.  


If the numbers (2n – 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers


In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms. 


Choose the correct alternative answer for  the following question .

 If for any A.P. d = 5 then t18 – t13 = .... 


The Sum of first five multiples of 3 is ______.


Choose the correct alternative answer for  the following question .

 In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....


Find the sum  (−5) + (−8)+ (−11) + ... + (−230) .


If the sum of first n terms of an A.P. is  \[\frac{1}{2}\] (3n2 + 7n), then find its nth term. Hence write its 20th term.

 
 

Write the sum of first n odd natural numbers.

 

Sum of n terms of the series  `sqrt2+sqrt8+sqrt18+sqrt32+....` is ______.


Find the sum of odd natural numbers from 1 to 101


The sum of all two digit odd numbers is ______.


Find the sum:

1 + (–2) + (–5) + (–8) + ... + (–236)


Find the sum:

`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms


Find the middle term of the AP. 95, 86, 77, ........, – 247.


Find the sum of all even numbers from 1 to 250.


If the first term of an A.P. is 5, the last term is 15 and the sum of first n terms is 30, then find the value of n.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×