The document explains arithmetic progression (AP), characterized by a constant difference between consecutive terms. It covers the formulation of the nth term, methods for determining terms and sums, and provides examples and practice problems related to AP. Key formulas include methods to compute the sum of finite arithmetic series and the general forms related to AP.

1. Arithmetic Progression

2.  Arithmetic Sequence is a sequence of
numbers such that the difference between
the consecutive terms is constant.
 For instance, the sequence 5, 7, 9, 11, 13,
15 … is an arithmetic progression with
common difference of 2.
 2,6,18,54(next term to the term is to
be obtained by multiplying by 3.

3. • A finite portion of an arithmetic progression
is called a finite arithmetic progression and
sometimes just called an arithmetic
progression. The sum of a finite arithmetic
progression is called an Arithmetic series.
• The behavior of the arithmetic progression
depends on the common difference d. If the
common difference is:
 Positive, the members (terms) will grow
towards positive infinity.
 Negative, the members (terms) will grow

4. Common Difference






an

an
1
4

6. Arithmetic Progression
 If the initial term of an arithmetic progression is a1 and the
common difference of successive members is d, then the
nth term of the sequence (an) is given by:

An= a1+(n-1)d
And in general
An= a1+ (n-m)d

7. The first term = a1 =a +0 d = a + (11)d second term = a = a + d = a + (2-1)d
The
2
The third term = a3 = a + 2d = a + (31)d fourth term = a = a + 3d = a + (4-1)d
The
4
------------------------------------------The nth term = an = a + (n------------------------------------------1)d

8. Example
Let a=2, d=2, n=12,find An
An=a+(n-1)d
=2+(12-1)2
=2+(11)2
=2+22
Therefore, An=24
Hence solved.

9. (i) 2, 6, 10, 14….
(i) Here first term a = 2,
find differences in the next terms
a2-a1 = 6 – 2 = 4
a3-a2 = 10 –6 = 4
a4-a3 = 14 – 10 = 4
Since the differences are common.
Hence the given terms are in A.P.

10. . Find 10th term of A.P. 12, 18, 24, 30.. …
Solution. Given A.P. is 12, 18, 24, 30..
First term is
a = 12
Common difference is d = 18- 12 = 6
nth term is
an = a + (n-1)d
Put n = 10,
a10 = 12 + (10-1)6
= 12 + 9 x 6
= 12 + 54
a10 = 66

11. .
Let A be the AM between a and b.
A, A, b are in AP
(A-a)=(b-A)
A=1/2(a+b)
AM between a and b
A=1/2(a+b)
Find the AM between 13 & 19
AM= ½(13+ 19) = 16

12. Some convenient methods to
determine the AP
• It is always convenient to make a choice of
i. 3 numbers in an AP as (a-d), a, (a+d);
ii. 4 numbers in an AP as (a-3d), (a-d), a,
(a+d),(a+3d);
iii. 5 numbers in an AP as (a-2d), (a-d), a, (a+d),
(a+2d).

13. Practice
Q---- The sum of three numbers in an AP is 21 & their
product is 231. Find the numbers.
Let the required Numbers be (a-d), a, (a+d)
Then (a-d)+ a+ (a+d)=21
 A=7
 And (a-d) a (a+d)=231
 A(a2-d2) =231
 Substituting the value of ‘a’
 7(49-d2)=231
 7d2= (343-231)= 112
 D2= 16
 D= 4

14. The sum of n terms, we find as,
Sum = n X [(first term + last term) / 2]
Now last term will be = a + (n-1) d
Therefore,

Sn = ½ n [ 2a + (n - 1)d ]
 It can also be written as

Sn = ½ n [ a + a n ]

15. Problem 1. Find the sum of 30 terms of given A.P.
12 + 20 + 28 + 36………
Solution : Given A.P. is 12 , 20, 28 , 36
Its first term is a = 12
Common difference is d = 20 – 12 = 8
The sum to n terms of an arithmetic progression
Sn = ½ n [ 2a + (n - 1)d ]
= ½ x 30 [ 2x 12 + (30-1)x 8]
= 15 [ 24 + 29 x8]
= 15[24 + 232]
= 15 x 246
= 3690

16. Practice
Q1: Find the sum of the following APs:
(i) 2, 7, 12, ……, to 10 terms
Q3: Given a = 2, d = 8, Sn = 90, find n
and an.
Question: 2. Find the sums given below:
(i) 7+10.5+14+…..+84

17. • Solution.
1) First term is
an = 500
a = 100 ,
9)
n - 1 = 80
10)
n = 80 + 1
n = 81
2) Common difference is d 11)
= 105 -100 = 5

Hence the no. of terms are 3) nth term is an = a + (n-81.
1)d
4) 500 = 100 + (n-1)5
5) 500 - 100 = 5(n – 1)
6) 400 = 5(n – 1)
7) 5(n – 1) = 400
8) n – 1 = 400/5

18. Practice
Question: 1. Find the sum of first 22 terms of an AP in which d = 7 and 22nd
term is 149.
Question: 2. Find the sum of first 51 terms of an AP whose second and third
terms are 14 and 18 respectively.

19. General Formulas of AP
• The general forms of an AP is a,(a+d), (a+2d),. .. , a +
( m - 1)d.
i. Nth term of the AP is Tn =a+(n-1)d.
ii. Nth term form the end ={l-(n-1)d}, where l is the
last term of the word.
iii. Sum of 1st n term of an AP is Sn=N/2{2a=(n-1)d}.
iv. Also Sn=n/2 (a+l)

20. MADE BY-: ,Anirudh Prasad, Ayush Gupta,
Chhavi Bansal&
Teena Kaushik
X- D
ROLL NO-6,10 , 12 & 33