- An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant. It can be represented by the first term (a), common difference (d), and last term (l). - The general formula for the nth term (Tn) of an AP is: Tn = a + (n-1)d - The sum (Sn) of the first n terms of an AP can be calculated as: Sn = (n/2)(a + l) - Inserting arithmetic means between two numbers a and b results in an AP where the common difference (d) is (b-a)/(n+1) and the inserted terms are: a + d, a + 2

1. Arithmetic Progression
(an Introduction to High school students)

2. Arithmetic Progression (A.P.)
• An arithmetic progression is a
sequence of numbers such that
the difference of any two
successive members is constant.
• Notations:
First term(T1) =a
Common difference =d
Last term(Tn) = l
• Let’s take a sequence
4, 7, 10 ,13, 16….
Can you find any pattern ????
First term (T1) = 4
Second term (T2)=7
Third term (T3) = 10
Fourth term(T4)=13
T2-T1 = 7-4 = 3
T3-T2 = 10-7 = 3
T4-T3 = 13-10 =3

3. Example-1: Check if the sequence 20, 10, 0, -10 is an AP ?
Solution:
what do we need to check ?
- common difference(d) between any
two successive terms should be
constant.
First term (T1) = 20
Second term(T2) =10
Third term(T3) = 0
Fourth term (T4) = -10
• Common difference between T1
and T2 (d) =T2-T1 = 10-20 = -10
• Common difference between T2
and T3 (d) =T3-T2 = 0-10 = -10
• Common difference between T3
and T4 (d) =T4-T3 = -10-0 = -10
• Since common difference (d) is -10
for all the terms, Hence Sequence
is an AP.

4. Example-2: Find the value of ‘k’ , if 5, k-2 , 11 are in A.P. ?
• Solution:
What do we know ???
T1= 5 , T2 =k-2 , T3=11
Next ?????
Common difference between any
two successive terms should be
constant in an A.P.
• difference between T2 and T1 =
difference between T3 and T2
(T2-T1) = (T3-T2)
(k-2-5) = (11-(k-2))
k-7 = 13-k
2k= 20
k=10

5. General form of an AP
First term (T1) = a
Second term (T2) = a+d
Third term (T3) = T2+d = a+2d
Forth term(T4) = T3+d = a+3d
nth term of an AP: Tn= a+(n-1)d
• So, General form of an AP will be:
a , a+d , a+2d , a+3d, …, a+(n-1)d
• If first term is ‘a’ and Common
difference is ‘d’

6. Example-4: Find the 23rd term of the sequence 2, 7, 12,…..
Solution-
check if it is an AP
d= T2 -T1 =5 , d= T3 -T2 = 5
Difference is constant so AP.
first term(T1) = 2
second term(T2) = 7 = 2+5
third term(T3) = 12 =2+5+5=2+2x5
23rd term(T23) = 2 + 22x5 = 112
• Directly using the formula if AP
Tn = a+(n-1)d
a=2 , d= 5 , n=23
T23 = 2+(23-1)x5 = 112

7. Sum of first n terms of an Arithmetic Progression
Example-5: Find the sum of first 100 natural numbers.
Solution:
first 100 naturals numbers = 1, 2, 3 ,4 ….., 98, 99 , 100
The sum of the first n terms of the series is denoted by Sn
Sn = T1 + T2 + T3 +… + Tn-1 + Tn
S100 = 1+2 +3 +4 +………..+99 +100 -----(i)
S100 = 100+99+98 +………..+2 +1 ------(ii) (Reversing the order of equation(i) )
2S100 = (1+100) +(2+99)+(3+98) +………+(99+2)+(100+1) (summation of (i) and (ii))
2S100 = 101 +101+101 +…………….+ 101+101
2S100 = 101 x 100 =10100
S100 = 10100/2 = 5050

8. Sum of first n terms of an Arithmetic Progression
• The sum of the first n terms of the series is denoted by Sn
• Sn = T1 + T2 + T3 +… + Tn-1 + Tn
Sn = (a) + (a+d) + (a+2d) + (a+3d) + …… + (a+(n-2)d)+(a+(n-1)d) --- (i)
Sn= (a+(n-1)d)+ (a+(n-2)d)+…………………+ (a+d) + (a) ----(ii)
by adding equation (i) and equation (ii)
2Sn = ( a+ a+(n-1)d) + (a+d+a+(n-2)d) +……..+ (a+(n-1)d +a)
2Sn= (2a+ (n-1)d) + (2a+(n-1)d) + ……….+ (2a +(n-1)d)
2Sn= (2a+(n-1)d) x (1+1+1….. n times)
2Sn = n (2a+(n-1)d)
Sn = (n/2)(2a+ (n-1)d)

9. Sum of first n terms of an Arithmetic Progression
• Sn = (n/2)(2a+(n-1)d)
• Sn = (n/2)( a+a+(n-1)d)
and as, Tn= a+(n-1)d = l
So , Sn =(n/2)(a+l)
where:
Sn= sum of first n terms of an AP
n- no of terms in an AP
a= first term of AP
l= last(nth) term of AP

10. Example-6: Find the sum of odd integers from 1 and 2001
• Sol: odd integers from 1 and 2001
are 1, 3,5,….1999, 2001
• Is this an AP ???
YES.. Because common difference is
constant.
• now , Sn = (n/2)(2a+(n-1)d)
or Sn =(n/2)(a+l)
• From the series: first term(a) =1 ,
common difference(d) = 3-1 =2
l=2001, no of terms (n ) = ??
Tn = a+(n-1)d
2001= 1+(n-1)2
n= 1001
• now sum of the sequence can be
find out using any of before
mentioned formula.
• Sn =(n/2)(a+l)
• S1001 = (1001/2) (1+2001)
• S1001= 1001x1001
• S1001 =1002001 (answer)

11. Arithmetic Mean (AM)
• If ‘A’ is the Arithmetic mean of two numbers ‘a’ and ‘b’ , then
a, A, b will form an AP.
A - a = b – A (common difference)
2A= (a + b)
A= (a + b)/2
Conclusion: Arithmetic mean is the average of two given numbers

12. Insertion of n Arithmetic means
• Let A1, A2, A3, ……,An be n
numbers between ‘a’ and ‘b’ such
that: a, A1, A2, A3,….,An , b forms
an AP.
• Here, b is (n+2)th term
(i.e. b= Tn+2)
Now Tn= a+(n-1)d
Tn+2= a+(n+2-1)d
b = a+(n+1)d
d= (b-a)/(n+1)
• A1 = a+d = a+ (b-a)/(n+1)
• A2 = a+2d = a+ 2(b-a)/(n+1)
• An = a+ nd = a+ n(b-a)/(n+1)

13. Example-6: Insert 3 numbers between 3 and 19 such
that resulting sequence is an AP.
Solution:
Let A1 , A2 , A3 are three numbers such that 3 , A1, A2, A3,19 are in A.P.
a= 3 , b= 19, n=3 , total no of terms(n+2)=5
Now to calculate ‘d’ ,
d=(b-a)/(n+1)
d=(19-3)/(3+1) =4
So, A1= a+d = 3+4 =7
A2= a+2d= 3+8 =11
A3= a+3d= 3+12=15
So, 3,7,11,15,19 will be the A.P.

14. .
THANK YOU