
01. Real Numbers
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02. Polynomials
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03. Linear Equation
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04. Quadratic Equation
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05. Arithmetic Progressions
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06. Some Applications of Trigonometry
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07. Coordinate Geometry
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08. Triangles
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09. Circles
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10. Areas Related to Circles
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11. Introduction to Trigonometry
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12. Surface Areas and Volumes
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13. Statistics
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14. Probability
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15. Construction
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Lecture15.7

Chapter Notes – Arithmetic Progressions
(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=2 which 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 4^{th} .
a=2, d=2 then we can find 10^{th} term by putting n=10 in above equation. So 10^{th} 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 10^{th} 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 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+(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 10^{th} term
Solution:
Given, 50,46,42,.....
Here , first term a=50,
Difference d=46−50=(−4)
And no of terms n=10
We know Sn=n2[2a+(n−1)d]
Sn=102[2(50)+(10−1)(−4)] ⇒5[100+(9)(−4)]
Sn=5[100−36]⇒5×64⇒320
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, an = 350 = l
And difference d = 9
We know, an=a+(n−1)d
350=17+(n−1)9
350=17+9n−9
9n=350−17+9⇒342
n=38
We know, sum of n terms
Sn=n2(a+l)
S38=382[17+350]⇒19×367⇒6973
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 nth terms, we replace n by (2n1) 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 mth terms.
Solution:
let a1, a2 be the 1^{st} terms and d1, d2 the common differences of the two given A.P’s. then the sums of their n terms are given by,
Sn1=n2{2a1+(n−1)d1} and Sn2=n2{2a2+(n−1)d2}
Sn1Sn2=n2{2a1+(n−1)d1}n2{2a2+(n−1)d2}
Sn1Sn2=2a1+(n−1)d12a2+(n−1)d2
It is given that Sn1Sn2=7n+14n+27
7n+14n+27=2a1+(n−1)d12a2+(n−1)d2………..(i)
To find ratio of the mth terms of the two given AP’s, we replace n by (2m−1) in equation (i). Therefore,
7(2m−1)+14(2m−1)+27=2a1+((2m−1)−1)d12a2+((2m−1)−1)d2
14m−68m+23=a1+(m−1)d1a2+(m−1)d2
Hence, the ratio of the mth terms of the two AP’s is 14m−68m+23
So as per rule if we replace n by (2m−1) we get ratio (14m−6):(8m+23)
(11) A sequence is an AP if and only if the sum of its n terms is of the form An2+Bn, where A, B are constants. In such a case the common difference is 2A.
For Example:
For the A.P
Sn=pn+q2n
Now S1=p×1+q(1)2
S1=p×q⇒T1=p+q and also S2=p×2+q(2)2
S2=2p+4q
We have T1+T2=2p+4q
Or T2=2p+4q−T1
T2=2p+4q−(p+q)⇒p+3q
Hence common difference = T2−T1
=p+3q−(p+q)=2q
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