This presentation explains the basics of Arithmetic Progression . It consists of introduction to arithmetic progression , first term and common difference along with some examples .
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Arithmetic Progression PART 1 ( INTRODUCTION )
1.
2. WHAT IS AN ARITHMETIC PROGRESSION ?
• Arithmetic progression is a sequence of numbers in which the
difference between every two consecutive terms is constant .
• For example :
1. 3,5,7,9,11,13,15,........ ( Here , the difference between every two
consecutive term is 2 )
2. 3,0,-3,-6,-9,-12,-15,-18. ( Here , the difference between every two
consecutive term is -3 )
3. 6,6,6,6,6,6,6,6,6,6,.......6. ( Here , the difference between every two
consecutive term is 0 )
A.P. in the 1st example is infinite as it does not end while A.P.'s in the
2nd and the 3rd examples are finite as they have a definite ending .
3. FIRST TERM AND COMMON DIFFERENCE OF
AN A.P.
1. First term of an A.P. is denoted by a or t1.
2. The difference between two consecutive terms of an A.P. is called the
common difference of the A.P. and it is denoted by d .
3. d of an A.P. is constant and it can be positive , negative or zero .
If an A.P. is t1, t2,t3,t4 ; then d= t2-t1 = t3-t2 = t4-t3
Thus , d = tn - tn-1
4. Is 5,8,16,32 an A.P. ?
Here , t1 = 5 , t2 = 8 , t3 = 16 , t4 = 32
t2-t1 = 8 - 5 = 3
t3-t2 = 16 - 8 = 8
t4-t3 = 32 - 16 = 16
The difference between every two
consecutive terms is not constant .
Thus, 5,8,16,32 is an A.P.
5. Is 0.5 ,1 , 1.5 , 2 an A.P. ? If yes , then what is
its first term and common difference ?
Here , t1 = 0.5 , t2 = 1 , t3 = 1.5 , t4 = 2
t2-t1 = 1- 0.5 =0.5
t3 -t2 = 1.5-1=0.5
t4 -t3 =2- 1.5=0.5
Here , the difference between every two consecutive terms is
constant.
Thus , 0.5 , 1 , 1.5 , 2 is an A.P.
In this A.P. , first term is 0.5 and common difference is 0.5 .
6. The first term and common difference of an
A.P. are 4 and -3 respectively . Find the A.P.
Let the A.P. be t1 , t2 , t3 , t4 , ..........
Here , t1 = 4 and d = -3
t2 = t1 + d = 4+(-3) = 4-3 = 1
t3 = t2 + d = 1+(-3) = 1-3 = -2
t4 = t3 + d = -2+(-3) = -2-3 = -5
Thus , the A.P. is 4,1,-2,-5,.........
