- Created By Admin
Class Members 447

- Respond To Some Unanswered Questions
- Arithmetic Progression
- set
- INTEGRATION

- Featured Questions and Posts
- simple equation
- Algebra
- binomial expansion
- change of subjects

- Study Classes
- › Accounting
- › Admissions/ Scholarships
- › Agricultural Science
- › Architecture
- › Biology
- › Business 101
- › Chemical Engineering
- › Chemistry
- › Civil Engineering
- › Computer
- › Current Affairs
- › Economics
- › English
- › General Science
- › Government
- › History
- › HTML/ Web Designing
- › Internet Business
- › Journalism
- › Law
- › Literature in English
- › Mass Communication
- › Mathematics
- › Mechanics
- › Microbiology
- › Motivation and Inspiration
- › Music
- › Nursing Science
- › PC Maintenance
- › PHP Programming
- › Physics
- › Political Science & Politics
- › Software Development

Example the following numbers form an arithmetic progression

2, 4, 6, 8, 10, 12, 14

the common difference d is 2 while the first term is 2

The common difference can be found by finding the difference between a term in the sequence and the preceding term eg from above 4 - 2 = 2

The general arithmetic progression from the above is

a, a + d, a + 2d, a + 3d, ...... where the first term, a and the common diference are arbitary numbers. The nth term of this progression is given by the formula

Tn = a + (n - 1)d.

Sum of Arithmetic Progression

To find the sum ofany arithmetic progression can be found by multiplying the sum of the first and last terms by half the number of terms. Eg the sum of the arithmetic progression of first ten natural numbers - 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 is (1 + 10) x (10 / 2) = 55

Therefore the formula for the first n terms of an arithmetic progression is

n/2[2a + (n - 1)d] 07 September 2010

Comment