# Arithmetic Progression - Class 10 : Notes

(1) A sequence is an arrangement of numbers or objects in a definite order.
For Example: 1, 8, 27, 64, 125,……
Above arrangement numbers are arranged in a definite order according to some rule.

(2) A sequence a1,a2,a3,....,an,.. is called an arithmetic progression, if there exists a constant d such that, a2a1=d,a3a2=d,a4a3=d,...,an+1an=d and so on. The constant d is called the common difference.
For Example: 2, 4, 6, 8,…. is a arithmetic progression because number are even natural numbers where a1=2,a2=4,a3=6,a4=8
42=2,64=2,86=d,...,an+1an=2

(3) If ‘a’ is the first term and 'd' the common difference of an AP,  then the A.P.  is a,a+d,a+2d,a+3d,a+4d....
For Example: If AP is 2, 4, 6, 8,…. Then first term a=2 and d=2
So, 2,2+4,2+2(2),2+3(2),a+4(2).....

(4) A sequence a1,a2,a3,....,an,.. is an AP, if an+1an is independent of n.
For Example: If sequence is 2, 4, 6, 8, …… an,….. so if we take an=16 so an+1=18 So an+1an=1816=2which is independent of n.

(5) A sequence a1,a2,a3,....,an,.. is an AP, if and only if its nth term an is a linear expression in n and In such a case the coefficient of n is the common difference.
For Example: A sequence 1, 4, 9, 16, 25,…. Is an AP. Suppose nth term an=81 which is a linear expression in n. which is n2.

(6) The nth term an, of an AP with first term ‘a’ and common difference ‘d’ is given by an=a+(n1)d
For Example: If want to find nth  term an in example given in 4th .
a=2d=2 then we can find 10th term by putting n=10 in above equation. So 10th term of  sequence is a10=2+(101)2=20

(7) Let there be an A.P with first term ‘a’ and common difference d. if there are m terms in the AP, then
nth term from the end = (mn+1)th
term from the beginning =a+(mn)d
Also, nth term from the end = Last term + (n1)(d)

l(n1)d, where l denotes the last term.
For Example: Determine the 10th term from the end of the A.P 4, 9, 14, …, 254.
l=254d=5
nth term from the end =l(101)d = l9d2549×5=209

(8) Various terms is an AP can be chosen in the following manner.

 Number of terms Terms Common difference 3 a−d,a,a+d d 4 a−3d,a−d,a+d,a+3d 2d 5 a−2d,a−d,a,a+d,a+2d d 6 a−5d,a−3d,a−d,a+d,a+3d,a+5d 2d

(9) The sum to n terms of an A.P with first term ‘a’ and common difference ‘d’ is given by Sn=n2{2a+(n1)d} Also, Sn=n2{a+l}, where l= last term = a+(n1)d
For Example: (i) 50, 46, 42, … find the sum of first 10th term
Solution:
Given, 50,46,42,.....
Here ,  first term ,
Difference
And no of terms
We know

Hence, Sum of 10 terms is 320.

(ii) First term is 17 and last term is 350 and d=9 so find total sum and find how many terms are there.
Solution:
Given, first term, a=17, last term,  = 350 =
And difference d = 9
We know,

We know, sum of n terms

Hence, number of terms is 38 and sum is 6973.

(10) If the ratio of the sums of n terms of two AP’s is given, then to find the ratio of their terms, we replace n by (2n-1) in the ratio of the sums of n terms.
For Example: The ratio of the sum of n terms of two AP’s is (7n+1):(4n+27). Find the ration of their terms.
Solution:
let  be the 1st terms and  the common differences of the two given A.P’s. then the sums of their n terms are given by,
and

It is given that
...........(i)

To find ratio of the  terms of the two given AP’s, we replace  by  in equation (i). Therefore,

Hence, the ratio of the  terms of the two AP’s is
So as per rule if we replace  by  we get ratio

(11) A sequence is an AP if and only if the sum of its n terms is of the form , where A,  B are constants.  In such a case the common difference is 2A.
For Example:
For the A.P

Now
and also

We have
Or

Hence common difference =

Notes for arithmetic progression chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

(1) A sequence is an arrangement of numbers or objects in a definite order.
For Example: 1, 8, 27, 64, 125,……
Above arrangement numbers are arranged in a definite order according to some rule.

(2) A sequence a1,a2,a3,....,an,.. is called an arithmetic progression, if there exists a constant d such that, a2−a1=d,a3−a2=d,a4−a3=d,...,an+1−an=d and so on. The constant d is called the common difference.
For Example: 2, 4, 6, 8,…. is a arithmetic progression because number are even natural numbers where a1=2,a2=4,a3=6,a4=8
4−2=2,6−4=2,8−6=d,...,an+1−an=2

(3) If ‘a’ is the first term and 'd' the common difference of an AP,  then the A.P.  is a,a+d,a+2d,a+3d,a+4d....
For Example: If AP is 2, 4, 6, 8,…. Then first term a=2 and d=2
So, 2,2+4,2+2(2),2+3(2),a+4(2).....

(4) A sequence a1,a2,a3,....,an,.. is an AP, if an+1−an is independent of n.
For Example: If sequence is 2, 4, 6, 8, …… an,….. so if we take an=16 so an+1=18 So an+1−an=18−16=2which is independent of n.

(5) A sequence a1,a2,a3,....,an,.. is an AP, if and only if its nth term an is a linear expression in n and In such a case the coefficient of n is the common difference.
For Example: A sequence 1, 4, 9, 16, 25,…. Is an AP. Suppose nth term an=81 which is a linear expression in n. which is n2.

(6) The nth term an, of an AP with first term ‘a’ and common difference ‘d’ is given by an=a+(n−1)d
For Example: If want to find nth  term an in example given in 4th .
a=2, d=2 then we can find 10th term by putting n=10 in above equation. So 10th term of  sequence is a10=2+(10−1)2=20

(7) Let there be an A.P with first term ‘a’ and common difference d. if there are m terms in the AP, then
nth term from the end = (m−n+1)th
term from the beginning =a+(m−n)d
Also, nth term from the end = Last term + (n−1)(−d)
= l−(n−1)d, where l denotes the last term.
For Example: Determine the 10th term from the end of the A.P 4, 9, 14, …, 254.
l=254, d=5
nth term from the end =l−(10−1)d = l−9d= 254−9×5=209

(8) Various terms is an AP can be chosen in the following manner.

Number of termsTermsCommon difference3a−d,a,a+dd4a−3d,a−d,a+d,a+3d2d5a−2d,a−d,a,a+d,a+2dd6a−5d,a−3d,a−d,a+d,a+3d,a+5d2d

(9) The sum to n terms of an A.P with first term ‘a’ and common difference ‘d’ is given by Sn=n2{2a+(n−1)d} Also, Sn=n2{a+l}, where l= last term = a+(n−1)d
For Example: (i) 50, 46, 42, … find the sum of first 10th term
Solution:
Given, 50,46,42,.....
Here ,  first term ,
Difference
And no of terms
We know

Hence, Sum of 10 terms is 320.

(ii) First term is 17 and last term is 350 and d=9 so find total sum and find how many terms are there.
Solution:
Given, first term, a=17, last term,  = 350 =
And difference d = 9
We know,

We know, sum of n terms

Hence, number of terms is 38 and sum is 6973.

(10) If the ratio of the sums of n terms of two AP’s is given, then to find the ratio of their terms, we replace n by (2n-1) in the ratio of the sums of n terms.
For Example: The ratio of the sum of n terms of two AP’s is (7n+1):(4n+27). Find the ration of their terms.
Solution:
let  be the 1st terms and  the common differences of the two given A.P’s. then the sums of their n terms are given by,
and

It is given that
...........(i)

To find ratio of the  terms of the two given AP’s, we replace  by  in equation (i). Therefore,

Hence, the ratio of the  terms of the two AP’s is
So as per rule if we replace  by  we get ratio

(11) A sequence is an AP if and only if the sum of its n terms is of the form , where A,  B are constants.  In such a case the common difference is 2A.
For Example:
For the A.P

Now
and also

We have
Or

Hence common difference =