Word Problems Involving Arithmetic Progression :

Example problems given in this section will be much useful for the students who would like to practice the concepts in arithmetic progression.

**Question 1 :**

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in each they are studying ,e.g, a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

**Solution :**

It can be observed that the number of trees planted by the student is in an AP

1, 2, 3, 4, 5, 6, ............... 12

To find total number of trees planted we have to find the sum of 12 terms

S_{n }= (n/2) [ 2 a + (n - 1) d]

S_{12} = (12/2) [ 2 (1) + (12 - 1) (1)]

= 6 [ 2 + 11 (1)]

= 6(13)

= 78

Total number of trees planted is 78.

**Question 2 :**

A spiral is made up of successive semicircles,with centres alternately at A and B starting with center at A of radii 0.5, 1.0 cm, 1.5 cm ,2.0 cm ......... as shown in figure . What is the total length of spiral made up of thirteen consecutive semicircles. (∏ = 22/7)

**Solution :**

Semi perimeter of a circle = ∏r

radius of first semicircle = 0.5 cm

semi perimeter of first circle = ∏ (1/2) = ∏/2

radius of second semicircle = 1 cm

semi perimeter of second circle = ∏ (1) = ∏

radius of third semicircle = 1.5 cm

semi perimeter of third circle = ∏ (3/2) = 3∏/2

(∏/2) + ∏ + (3∏/2) + ...........................

S_{n}= (n/2) [ 2a + (n - 1)d ]

S_{13} = (13/2) [2(∏/2)) + (13-1) ((∏/2))]

= (13/2) [ ∏ + 6 ∏]

= (13/2) [7∏]

= 91∏/2

= (91/2) x (22/7)

= 143

Hence the length of spiral of thirteen consecutive semicircles will be 143 cm.

**Question 3 :**

A sum of Rs 700 is to be used to seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than the preceding prize, find the value of each prizes.

**Solution :**

x + (x - 20) + (x - 40) + ............... = 700

S_{n} = 700

a = x, d = x - 20 - x = -20, n = 7

S_{n }= (n/2) [ 2 a + (n - 1) d]

S_{7} = (7/2) [2(x) + (7 - 1) (-20)]

700 = (7/2) [2x + 6(-20)]

700 = (7/2) [2x - 120]

(700 x 2)/7 = 2 x - 120

200 = 2x - 120

2x = 200 + 120

2 x = 320

x = 320/2 = 160

The first prize amount Rs 160

Second prize = Rs 140

Third prize = Rs 120

Fourth prize = Rs 100

Fifth prize = Rs 80

sixth prize = Rs 60

Seventh prize = Rs 40

After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems involving arithmetic progression.

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