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?
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
Sn = (n/2) [ 2 a + (n - 1) d]
S12 = (12/2) [ 2 (1) + (12 - 1) (1)]
= 6 [ 2 + 11 (1)]
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)
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) + ...........................
Sn= (n/2) [ 2a + (n - 1)d ]
S13 = (13/2) [2(∏/2)) + (13-1) ((∏/2))]
= (13/2) [ ∏ + 6 ∏]
= (13/2) [7∏]
= (91/2) x (22/7)
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.
x + (x - 20) + (x - 40) + ............... = 700
Sn = 700
a = x, d = x - 20 - x = -20, n = 7
Sn = (n/2) [ 2 a + (n - 1) d]
S7 = (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.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
APTITUDE TESTS ONLINE
ACT MATH ONLINE TEST
TRANSFORMATIONS OF FUNCTIONS
ORDER OF OPERATIONS
Converting customary units worksheet
Customary units worksheet
Integers and absolute value worksheets
Nature of the roots of a quadratic equation worksheets
Trigonometry heights and distances
MATH FOR KIDS
Word problems on linear equations
Trigonometry word problems
Word problems on mixed fractrions
Converting repeating decimals in to fractions