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Session MPTCP04. Sequences and Series. Session Objectives. Finite and infinite sequences Arithmetic Progression (A.P.) - definition, n th term Sum of n terms of an A.P. Arithmetic Mean (A.M.) and insertion of n A.M.s between two given numbers.
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Session MPTCP04 Sequences and Series
Session Objectives • Finite and infinite sequences • Arithmetic Progression (A.P.) - definition, nth term • Sum of n terms of an A.P. • Arithmetic Mean (A.M.) and insertion of n A.M.s between two given numbers. • Geometric Progression (G.P.) - definition, nth term • Sum of n terms of a G.P.
Sequence – a Definition _I001 A sequence is a function whose domain is the set N of natural numbers. a1, a2, a3, . . ., an, . . .
Finite and Infinite Sequences _I001 Finite sequence : a a1, a2, a3, . . ., an Infinite sequence : a a1, a2, a3, . . ., an, . . .
Series – a Definition _I001 If a1, a2, a3, . . ., an, . . . is a sequence, the expression a1+a2+a3+ . . . +an+ . . . is called a series.
Arithmetic Progression _I002 A sequence is called an arithmetic progression (A.P.) if the difference between any term and the previous term is constant. The constant difference, generally denoted by d is called the common difference. a1 = a a2 = a+d a3 = a+d+d = a+2d a4 = a+d+d+d = a+3d First term General Term an = a+d+d+d+... = a+(n-1)d
Is a Given Sequence an A.P.? _I002 Algorithm to determine whether a given sequence is an A.P. : Step I Obtain general term an Step II Determine an+1 by replacing n by n+1 in the general term Step III Find an+1-an. If this is independent of n, the given sequence is an A.P.
Problem Solving Tip _I002 Choose Well!!!! # Terms Common diff. 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
Illustrative problem _I002 Q. If sum of three numbers in A.P. is 45, and the second number is thrice the first number, find the three numbers. A. Let the numbers be a-d, a, a+d Given that (a-d)+a+(a+d) = 45 3a = 45 a = 15 Also, a = 3(a-d) 3d = 30 d = 10 the three numbers are 5, 15, 25
Important Properties of A.P.s _I002 If A a, a+d, . . ., a+(n-1)d adding constant k to each term, we get, A’ a+k, a+d+k, . . ., a+(n-1)d+k A’ is also an A.P. with the same common difference.
Important Properties of A.P.s _I002 If A a, a+d, . . ., a+(n-1)d multiplying each term by non-zero constant k, A’ ak, ak+dk, . . ., ak+(n-1)dk A’ is also an A.P. with common difference dk
Important Properties of A.P.s _I002 ak+an-(k-1) = a1+an k = 2, 3, 4, . . . (n-1) Example : Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Important Properties of A.P.s _I002 ak+an-(k-1) = a1+an k = 2, 3, 4, . . . (n-1) Example : Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 a3+a8 = 22
Important Properties of A.P.S _I002 ak+an-(k-1) = a1+an k = 2, 3, 4, . . . (n-1) Example : Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 a3+a8 = 22 = a5+a6 = 22
Important Properties of A.P.s _I002 ak+an-(k-1) = a1+an k = 2, 3, 4, . . . (n-1) Example : Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 a3+a8 = 22 = a5+a6 = 22 = a1+a10 = 22
Important Properties of A.P.s _I002 a, b, c are in A.P. 2b = a+c
Important Properties of A.P.s _I002 A sequence is an A.P. an = An+B, A, B are constants. A is the common difference. Proof : an = a+(n-1)d or, an = dn+(a-d) or, an = An+B, where A is the common difference
Important Properties of A.P.s _I002 If A a, a+d, . . ., a+(n-1)d take every third term, A’ a, a+3d, a+6d, . . . . . . . . . . A’ is also an A.P.
Sum of n Terms of an A.P. _I003 Sn = a1 +(a1+d)+ . . .+{a1+(n-2)d}+{a1+(n-1)d} Also, Sn = {a1+(n-1)d}+{a1+(n-2)d}+{a1+d}+. . .+a1 Adding, 2Sn = n{2a1+(n-1)d}
Sum of n Terms of an A.P. _I003 This can also be written as :
We know that, Rearranging, Property of Sum of n Terms of an A.P. _I003 A sequence is an A.P. Sn = An2+Bn, where A, B are constants. 2A is the common difference. Or, Sn = An2+Bn.
Arithmetic Mean _I004 A is the A.M. of two numbers a and b a, A, and b are in A.P. A-a = b-A 2A = a+b
-4 -2 0 2 4 6 -6 A1 A2 A3 A4 A5 b a Arithmetic Mean – a Definition _I004 If n terms A1, A2, A3, . . . An are inserted between two numbers a and b such that a, A1, A2, A3, . . . , An, b form an A.P., then A1, A2, A3, . . . , An are called arithmetic means (A.M.s) of a and b.
-4 -2 0 2 4 6 -6 A1 A2 A3 A4 A5 b a Arithmetic Mean – Common Difference _I004 Let n A.M.s be inserted between two numbers a and b The A.P. thus formed will have (n+2) terms. Let the common difference be d Now b = a+(n+2-1)d = a+(n+1)d
Property of A.M.s _I004 Let n A.M.s A1, A2, A3, . . ., An be inserted between a and b. Then,
A. Let the required A.M.s be A1, A2 and A3. Common difference d = Illustrative Problem _I004 Q. Insert 3 A.M.s between -4 and 3
Geometric Progression _I005 Consider a family where every female of each generation has exactly 2 daughters. It is then possible to determine the number of females in each generation if the generation number is known. 1st Generation 1 female 2nd Generation 2 females 3rd Generation 4 females Such a progression is a Geometric Progression (G.P.)
Geometric Progression _I005 A sequence is called a geometric progression (G.P.) if the ratio between any term and the previous term is constant. The constant ratio, generally denoted by r is called the common ratio. a1 = a a2 = ar a3 = ar2 a4 = ar3 First term General Term an = ar(n-1)
Problem Solving Tip _I005 Choose Well!!!! # Terms Common ratio 3 a/r, a, ar r 4 a/r3, a/r, ar, ar3 r2 5 a/r2, a/r, a, ar, ar2 r 6 a/r5, a/r3, a/r, ar, ar3, ar5 r2
Important Properties of G.P.s _I005 If G a, ar, ar2, . . ., arn-1 multiplying each term by non-zero constant k, G’ ka, kar, kar2, . . ., karn-1 G’ is also a G.P. with the same common ratio.
Important Properties of G.P.s _I005 If G a, ar, ar2, . . ., arn-1 taking reciprocal of each term, G’ is also a G.P. with a reciprocal common ratio.
Important Properties of G.P.s _I005 If G a, ar, ar2, . . ., arn-1 raising each term to power k, G’ ak, akrk, akr2k, . . ., akr(n-1)k G’ is also a G.P. with common ratio rk.
Important Properties of G.P.s _I005 akan-(k-1) = a1an k = 2, 3, 4, . . . (n-1) Example : Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512
Important Properties of G.P.s _I005 akan-(k-1) = a1an k = 2, 3, 4, . . . (n-1) Example : Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 a3a8 = 512
Important Properties of G.P.s _I005 akan-(k-1) = a1an k = 2, 3, 4, . . . (n-1) Example : Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 a3a8 = 512 = a5a6 = 512
Important Properties of G.P.s _I005 akan-(k-1) = a1an k = 2, 3, 4, . . . (n-1) Example : Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 a3a8 = 512 = a5a6 = 512 = a1a10 = 512
Important Properties of G.P.s _I005 a, b, c are in G.P. b2 = ac
Important Properties of G.P.s _I005 If G a, ar, ar2, . . ., arn-1 take every third term, G’ a, ar3, ar6, . . . G’ is also a G.P.
Important Properties of G.P.s _I005 a1, a2, a3, . . . , an is a G.P. of positive terms loga1, loga2, loga3, . . . logan is an A.P.
Sum of n Terms of a G.P. _I006 Sn = a+ar+ar2+ar3+ . . .+ar(n-1) ………(i) Multiplying by r, we get, rSn = ar+ar2+ar3+ . . .+ar(n-1)+arn ……...(ii) Subtracting (i) from (ii), (r-1)Sn = a(rn-1)
Class Exercise Q1. _I002 Q. If log 2, log (2x-1) and log (2x+3) are in A.P., find x.
Class Exercise Q1. _I002 Q. If log 2, log (2x-1) and log (2x+3) are in A.P., find x. A. Given that log(2x-1)-log2 = log(2x+3)-log(2x-1)
Class Exercise Q2. _I002 Q. Show that there is no infinite A.P. which consists only of distinct primes.
Class Exercise Q2. _I002 Q. Show that there is no infinite A.P. which consists only of distinct primes. A. Let, if possible, there be an A.P. consisting only of distinct primes : a1, a2, a3, . . ., an, . . . an = a1+(n-1)d Thus, (a1+1)th term is a multiple of a1. Thus, no such A.P. is possible. Q.E.D.
Q. , where Sn denotes the sum of the first n terms of an A.P., then common difference is : (a) P+Q (b) 2P+3Q (c) 2Q (d) Q (J.E.E. West Bengal 1994) Class Exercise Q3. _I003
Q. , where Sn denotes the sum of the first n terms of an A.P., then common difference is : (a) P+Q (b) 2P+3Q (c) 2Q (d) Q (J.E.E. West Bengal 1994) Class Exercise Q3. _I003 A. an = Sn - Sn-1 Ans : (d).
Class Exercise Q4. _I003 Q. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
Class Exercise Q4. _I003 Q. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms? A. Given that, a12 = a1+11d = -13 . . . (i) S4 = 2(2a1+3d) = 24 . . . (ii) Solving (i) and (ii) simultaneously, we get, a1 = 9, d = -2 S10 = 5(2a1+9d) = 5(18-18) = 0
Q. Find the value of n so that be an A.M. between a and b (a, b are positive). Class Exercise Q5. _I004
Q. Find the value of n so that be an A.M. between a and b (a, b are positive). Class Exercise Q5. _I004 A. Given that, Dividing throughout by bn+1, we get,