Arithmetic progression For class 10. In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant

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