sequence and series

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SEQUENCE & PROGRESSIONMaximum no. of lecture allowed : 9 for Bull's Eye ; 8 for ACME ; 6 for 13th

1ST LECTURESyllabus in IIT JEE : Arithmetic, geometric and harmonic progressions, arithmetic,geometric and harmonic means, sum of finite arithmetic and geometric progressions,infinite geometric series, sum of square and cubes of the first n natural numbers.

1. INTRODUCTION :A sequence is a set of terms which may be algebraic, real or complex numbers, writtenaccording to definite rule and the series thus formed is called a progression.e.g. 0, 1, 7, 26.............. (rule is n3 1)

1, 4, 7, 10 .............2, 4, 6, 8, ............... etc.

Note : Minimum number of terms in a sequence should be 3.2. SHORT STORY :

Leading to historical development and origin of the chapter - Fredric Karl Gauss.

3(a) ARITHMETIC PROGRESSION :It is a sequence whose terms increase or decrease by a fixed number. Fixed number iscalled the common difference. If 'a' is the first term and 'd' is the common difference,then the standard appearance of an A.P. is

a + (a + d) + (a + 2d) + ......... + (a + 1n d )and nth or last term is given by

Tn = a + (n 1)dNote : If d > 0 increasing A.P. If d < 0 decreasing A.P. If d = 0 all the terms remain same

EXAMPLES :(i) If 6th and 11th term of an A.P. are respectively 17 and 32. Find the 20th term. [Ans. 59](ii) In an A.P. if tp = q and tq = p then find the rth term. [Ans. p + q r](iii) In an A.P. if a2 + a5 a3 = 10 and a2 + a9 = 17 then find the 1st term and the commondifference. [a1= 13 and d = 1](iv) If pth, qth and rth term of an A.P. are respectively a, b, and c then prove that

a(q r) + b(r p) + (p q) = 03(b) Sum of n terms of an A.P.

Sn = a + (a + d) + (a + 2d) + .............. + (a + 1n d )Sn = (a + 1n d ) + (a + 2n d ) + (a + d) +.................+ a2Sn = n [2a + (n 1)d]Sn = 2

n [2a + 1n d] or 2n (a + l) where l = a + 1n d]

Remember that : (i) sum of first n natural number is 2)1n(n

(ii) sum of first n odd natural number is n2(iii) sum of first n even natural number is n(n + 1)

TEACHING NOTES


3(c) HIGHLIGHTS ABOUTAN A.P.(i) If each term of an A.P. is increased, decreased, multiplied or divided by the same non

zero number, then the resulting sequence is also an AP.(ii) Three numbers in AP can be taken as a d , a , a + d ; four numbers in AP can be taken

as a 3d, a d, a + d, a + 3d ; five numbers in AP are a 2d, a d , a, a + d, a + 2d &six terms in AP are a 5d, a 3d, a d, a + d, a + 3d, a + 5d etc.

(iii) The common difference can be zero, positive or negative.(iv) The sum of the two terms of an AP equidistant from the beginning & end is constant

and equal to the sum of first & last terms.(v) If the number of terms in an A.P. is even then take it as 2n and if odd then take it as (2n+1)(vi) For any series, Tn = Sn Sn 1. In a series if Sn is a quadratic function of n or Tn is alinear function of n, then the series is an A.P.(vii) If a, b, c are in A.P. 2b = a + c.EXAMPLES :(i) The first term of an A.P. is 5, the last is 45, and the sum 400. Find the number of terms

and the common difference. [Ans. n = 16, d = 322 ]

(ii) The sum of first 3 terms of an A.P. is 27 and the sum of their squares is 293. Find Sn.[Ans. Sn = 2

n (5n + 3) or 2n (33 5n)]

(iii) In an A.P. if tm = n1 and tn = m

1 then show that Smn = 21 (mn + 1)

(iv) How many terms of the sequence,20 + 3

119 + 3218 + ........ must be taken so that there sum is 300. Explain the reason of

double answer [Ans. 36 or 25](v) The sum of n terms of two A.P.'s are in the ratio of 7n + 1 : 4n + 27, find the ratio of

their 11th terms [Ans. 34 ]

(vi) If S1, S2, S3, ...... Sp are the sums of n terms of 'p' arithmetic series whose first termsare 1, 2, 3, 4........ and whose common difference are 1, 3, 5, 7, ......; prove thatS1 + S2 + S3 + ....... + Sp = )1np(2

np

[Hint: S1 = 2n [2 + (n 1)] = 2

)1n(n

S2 = 2n [4 + 1n 3] = 2

)1n3(n

Sp = 2n [2p + 1n (2p 1)] = 2

n

2

1n)1p2(

S1 + S2 + ..... + Sp = 2n [n + 3n + 5n + ...... + (2p 1)n + p]

= 2n [n(1 + 3 + 5 + ..... + 2p 1) + p] = 2

n [np2 + p] = 2np (np+1) ]


(vii) In an A.P. Sp = q = and Sq = p then show that Sp+q = (p + q)(viii) The number of terms in an A.P. is even, the sum of the odd terms is 24, of the even

terms is 30, and the last term exceeds the first term by 2110 . Find the number of terms.

(ix) Find the sum of all integers between 1 to 100 which are divisible by 2 or 3.(x) Find the nature and nth term of the sequence whose sum to n terms is 5n2 + 2n + 4.

[Ans. Tn = 10n 3 (an A.P.) 7, 17, 27, 37 .........]Asking : S = 1002 992 + 982 972 + 962 952 + ........ + 22 12 = 5050.Note :

(1) If a, b, c are in A.P. then prove that(a) b + c ; c + a ; a + b are also in A.P.[Hint: b + c = a + d + a + 2d etc, c + a = a + 3d + a = 3a + 3d etc.](b) (b + c)2 a2 ; (c + a)2 b2 ; (a + b)2 c2 are also in A.P.[Hint: b + c a, c + a b, a + b c are in A.P.

b + c a = a + d + a + 2d a = a + 3d etc. ](2) If a2, b2, c2 are in A.P. then prove that

(a) cb1 , ac

1 , ba

1 are in A.P.. (b) cb

a , ac

b , ba

c are in A.P..

[Sol. a2, b2, c2 are in A.P. (given)2(a) a2 + ab + bc + ca, b2 + ab + bc + ca, c2 + ab + bc + ca are in A.P.

(a + b)(a + c), (b + c)(b + a), (c + a)(c + b) are in A.P.or cb

1 , ac

1 , ba

1 are in A.P..

2(b) TPT cba , ac

b , ba

c are in A.P..

or TPT cba + 1, ac

b + 1, ba

c + 1 are in A.P..

or TPT cbcba

, ac

cba , ba

cba are in A.P..

or TPT cb1 , ac

1 , ba

1 are in A.P. ]

Home Work after 1st lecture : Ex. IV (a) and IV (b) except mean (Hall & Knight)2ND LECTURE

3(d) ARITHMETIC MEAN :Definition : When three quantities are in A.P. then the middle one is called theArithmeticMean of the other two.e.g. a, b, c are in A.P. then 'b' is the arithmetic mean between 'a' and 'c' and a + c = 2b.It is to be noted that between two given quantities it is always possible to insert anynumber of terms such that the whole series thus formed shall be in A.P. and the termsthus inserted are called the arithmetic means.To insert 'n' AM's between a and b.Let A1, A2, A3 ........ An are the n means between a and b.Hence a, A1A2, ........ Anb is an A.P. and b is the (n + 2)th terms.Hence b = a + (n + 1)d d = 1n

ab


Now A1 = a + dA2 = a + 2dAn = a + nd

n

1A

i i = na + (1 + 2 + 3 + .... + n)d

= na + d2)1n(n = na + 1n

ab2)1n(n

= ]aba2[2n = n

2ba = na

Hence the sum of n AM's inserted between a and b is equal to n times a single AMbetween them.

Asking : If 101 means are inserted between 1 and 99 then their sum = 2)991()101( =5050.

EXAMPLES :(i) Insert 20 AM's between 4 and 67.(ii) If p arithmetic means are inserted between 5 and 41 so that the ratio

1p3

AA

= 5

2 then

find the value of p. [Ans. p = 11](iii)109\1 A number sequence a1, a2, a3 ....... an is such thata1 = 0 ; | a2 | = | a1 + 1 | ; | a3 | = | a2 + 1 | ...... | an | = | an 1 + 1 |.

Prove that the arithmetic mean of a1, a2, ....... an is not less than 21 .

[Sol. Given a1 = 0 ; | a2 | = | a1 + 1 | ; | a3 | = | a2 + 1 | ........... | an | = | an 1 + 1 |also let an + 1 = | an + 1 |now squaring 21a = 0

22a = 1a2a 121

1a2aa 22223 1a2aa 32324

1a2aa n2n2 1n

adding 2

1n2n

22

21 aa.....aa = n)a.....aa(2a.....aa n212n2221

or n)a.....aa(2a n212 1n or 2 (a1 + a2 + ...... + an) = na2 1n hence, 2 (a1 + a2 + ...... + an) n n

a....aa n21 21 ]


GENERAL ILLUSTRATION ON A.P.(i)(a)126/1 If log3 2, log3 (2x 5) &

2

72log x3 are in AP, determine x.[Ans: x = 3] [IIT90, 4]

[Sol. log ( ) log log log ( )3 3 3 32 5 2 2 72 2 5x x x FH IK ; log log3 32 52

2 722 5

x x

xFHG

IKJ

FHGGG

IKJJJ

2 52

2 722 5

x x

x

; 2 5 2 7

2 1x x e j2 10 2 25 2 2 72x x x . . ; put 2x = yy y y or2 12 32 0 8 4 ; hence x = 3 (as x = 2 is rejected) ]

(b) Solve the equation x1........x

2xx1x = 3 [Ans. x = 7]

[Sol. x1 [(x 1) + (x 2) + (x 3) + ... + (x (x 1))] = 3x(x 1) (1 + 2 + 3 + ... + (x 1)) = 3xx(x 1) 2

x)1x( = 3x

x(x 1) 21 = 3x

x2 x = 6xx(x 7) = 0 ; x 0 x = 7 ]

(ii)(a)132/1 Find the condition that the roots of the equation x3 px2 + qx r = 0 may be in A.P.and hence solve the equation x3 12x2 + 39x 28 =0.[Hint: Let the roots be a d, a and a + d 3a = p

a = 3p must satisfy the given equation

0r3pq9

p.p27p 23 2p3 9pq + 27r = 0 ....(1)(This is the required condition)now given, x3 12x2 + 39x 28 = 0 ....(2)here p = 12; q = 39 and r = 28 which satisfy (1)hence roots of equation (2) are also in A.P. a = 3

123p = 4

also a(a d) (a + d) = r = 28a2 d2 = 7d2 = 9 d = + 3

Hence the roots are 1, 4, 7 Ans. ](b) If the first 3 terms of an increasing A.P. are the roots of the cubic

4x3 24x2 + 23x + 18 = 0, then find Sn. [Ans. (5n 7) 4n ]

[Hint: series is 21 , 2, 2

9 , ..........]


(iii)151/1 If the sum of the roots of the equation ax + bx + c = 0 is equal to the sum of thesquares of their reciprocals, then show that bc, ca, ab are in AP.[Sol. ax2 + bx + c = 0 ; + = a

b ; = ac

Given + = 2211 + = 22

2 2)(

ab

22

ac = a

c2ab22 2

2

ac = a

b + bc2

bc2 = ab2 + 2ca2bc2 + ab2 = 2ca2 bc2, ca2, ab2 in AP ]

(iv) Given a1, a2, a3.....an in A.P. Prove that

n1aa1 +

1n2aa1

+

1n2aa1

+ ...... +

1naa1 =

n2121 a

1.....a1

a1

aa2

(v) Prove that 2 , 3 , 5 can not be the terms of an A.P. (not necessarily adjacent)[Hint: Let 2 , 3 , 5 are pth , qth and rth terms.

Hence 2 = a + (p 1)d ; 3 = a + (q 1)d ; 5 = a + (r 1)d

2335

= pq

qr ; RHS is rational and LHS is irrational ]

Home Work after 2nd lecture : Ex. IV (a) and IV (b) including mean., (Hall & Knight)3RD LECTURE

4(a) GEOMETRICAL PROGRESSION :Definition : In a sequence if each term (except the first non zero term) bears the sameconstant ratio with its immediately preceding term the series is called a G.P. and theconstant ratio is called the common ratio.Standard appearance of a G.P. is

a + ar + ar2 + ar3 + ..........+ arn 1 , where nth term is Tn = arn 1EXAMPLES :(i) In a G.P. if t3 = 2 and t6 = 4

1 find t10. [Ans. 641 ]

(ii) If pth, qth and rth terms of a G.P. are x, y and z respectively hen prove thatxq r yr p zp q = 1

4(b) Sum of n terms of a G.P.S = a + ar + ar2 + .......... + arn 1Sr= + ar + ar2 + ...................... + arn

subtract S(1 r) = a arn = a (1 rn)S = r1

)r1(a n , where r 1, if r = 1 then S = na

If | r | < 1 and n then rn 0 and in this case geometric series will be summableupto infinity and its sum is given by

S = r1a


4(c) IMPORTANT POINTS TO REMEMBER(i) If each term of a GP be multiplied or divided by the same non-zero quantity, the

resulting sequence is also a GP.(ii) Any 3 consecutive terms of a GP can be taken as a/r, a, ar ; any 4 consecutive terms

of a GP can be taken as a/r3, a/r, ar, ar3 & so on.(iii) If a, b, c are in GP b2 = ac.EXAMPLES :(i)(a) The sum of first 3 consecutive terms of a G.P. is 19 and their product is 216. Find Sn ,also compute s if it exist.[Sol. Let the three numbers in GP are r

a , a, ar

a

r1r1 = 19 ...(1) r

a 1ar = 216 ....(2) a = 6now from (1) r

1 + 1 + r = 619 r + r

1 = 613 r2 6

r13 + 1 = 06r2 13r + 6 = 0 6r2 9r 4r + 6 = 0 3r(2r 3) 2(2r 3) = 0r = 3

2 or r = 23 .

Hence GP's are (i) 4, 6, 9, 227 , ... (ii) 9, 6, 4, 3

8 , ...

In 1st case Sn = 123

1234

n

= 8

123 n ;

In 2nd case Sn =3213219

n

; Sn = 27

n

321

S exists if | r | < 1. Hence in 1st case S does not exist and in 2nd case S = 27 ] (b) In a G.P. with an

a1 + a2 + a3 = 13 and 232221 aaa = 91. Find Sn.(ii) The sum of an infinite number of terms of a G.P. is 15 and the sum of their squares is

45. Find the series. [Ans. 5, 1310 , 9

20 , .........]

(iii) Evaluate

n1r

n

1ssr

rs 32 (an elegant way of expressing a G.P.)

where rs =

srif1srif0


(iv) (a) Use infinite series to compute the rational number corresponding to 234.0(b) Find the sum S = 9 + 99 + 999 + ...... + 9......999 n times(c) S = 0.9 + 0.99 + 0.999 + ....... up to n times.

(v) Find the four successive terms of a G.P. of which the 2nd term is smaller than the firstby 35 and the 3rd term is larger than the 4th by 560. [Ans. 7, 28, 112, 448]

[Hint. a, ar, ar2, ar3 a ar = 35 ....(1)ar2 ar3 = 560 .....(2) ]

(vi) If the pth, qth, rth, sth terms of an A.P. are in G.P.show that p q, q r , r s are in G.P. [T/S, Ex-1]

[Sol. Tp, Tq, Tr and Ts are in A.P. d)1q(a

d)1p(a = d)1r(a

d)1q(a = d)1s(a

d)1r(a

each of these ratios= d)1r(ad)1q(a

d)1q(ad)1p(a = d)1s(ad)1r(a

d)1r(ad)1q(a

= rqqp

= sr

rq

Hence p q, q r, r s are in G.P. ](vii) If S1, S2, S3,......Sp are the sums of infinite G.P. whose first terms are 1, 2, 3, .......p,

and whose common ratios are 21 , 3

1 , 41 , ........, 1p

1 respectively..

Prove that S1 + S2 + S3 + .......Sp = 2p (p + 3).

Home Work after 3rd lecture : Ex. 5(a) except mean (Hall & Knight)4TH LECTURE

4(d) GEOMETRICAL MEANDefinition : If a, b, c are three positive number in G.P. then b is called the geometricalmean between a and c and b2 = ac. If a and b are two + ve real and G is the G.M.between them, then

G2 = abTo insert 'n' GM's between a and bLet a and b are two positive numbers are G1, G2, .........Gn are 'n' GM's thena G1 G2 ...........Gn b is a G.P. with 'b' as its (n+2)th term.Hence b = arn + 1

r = 1n1

ab

Now G1 = ar, G2 = ar2, ..........Gn = arn

hence

n

1G

ii = an r1 + 2 + .... + n = an 2

)1n(nr

=

2)1n(n

1n1

naba

= 2n2nn

aba = an/2 bn/2 = nab = Gn

where G is the angle GM between a and b.Hence product of n GM's inserted between of a and b is equal to the nth power of asingle GM between them. It may be noted that between two positive numbers AM GM


EXAMPLES :(i) Insert 4 GM's between 5 and 160.(ii) If AM between a and b is 15 and GM between a and b is 9. Find the number.(iii) If sum of two numbers a and b is n times their GM then show that

a : b = 4nn 2 : 4nn 2 [Sol. a + b = n ab b

a + ab = n

let ba = y ; y + y

1 = n y2 ny + 1 = 0 y = 24nn 2 ]

(iv) If a, b, c are in G.P. and x, y are respectively the AM's between a, b and b, c then prove that

x1 + y

1 = b2 and x

a + yc = 2

[Sol. a = a ; b = ar ; c = ar2also a, x, b in A.P and b y c in A.P.

2x = a + b = a(1 + r) ....(1)2y = b + c = ar(1 + r) ....(2)

now x1 + y

1 = )r1(a2 + )r1(ar

2 = )r1(ar

)r1(2 = ar

2 = b2

again xa + y

c = r12 + )r1(ar

2)ar( 2 [from 1 and 2]

= r12 + )r1(

r2 = 2 ]

(v) If x > 0, y > 0, z > 0 then prove that (x + y)(y + z)(z + x) 8xyz[Hint: x + y yx2 etc.]( v i ) P r o v e t h a t a ABC is equilateral if and only

tanA + tanB + tanC = 33[Hint: Case-1 : Let ABC is equilateral then tanA + tanB + tanC = 33 is obviously true

Case-2 : Let tanA + tanB + tanC = 33 be true

3CtanBtanAtan = 3

AM of tanA, tanB, tanC = 3now GM of tanA, tanB, tanC = (tanA tanB tanC)1/3 = 3133 = 31233 = 3Hence AM = GM ; tanA = tan B = tan C is equilateral ]

(vii) If a + b + c = 3 and a, b, c are +ve then prove that a2b3c2 7410

723

[Sol. applying AM GM in seven numbers 2a , 2

a , 3b , 3

b , 3b , 2

c , 2c

7cba

7134232

32cba

;

7

73

34232

32cba ; a2b3c2 7

104

732 ]

H.W. after 4th lecture : Ex. 5(a) including means + Ex-1, T/S 1 to 15(Hall & Knight)


5TH LECTURE5 ARITHMETIC GEOMETRIC PROGRESSION (AGP) :

Standard appearance of an AGP isS = a + (a + d)r + (a + 2d)r2 + (a + 3d)r3 + ..........

Here each term is the product of corresponding terms in an arithmetic and geometric series.LetS = a + (a + d)r + (a + 2d)r2 + (a + 3d)r3 + .......... + d1na rn 1Sr = + ar + (a + d)r2 + ...............................+ d2na rn 1 + d1na rnNow subtract and get the expression for S and S as the case may beEXAMPLES :

(i) If | x | < 1 then compute the sum(a) 1 + 2x + 3x2 + 4x3 + ........... [Ans. 2)x1(

1 ]

(b) 1 + 3x + 6x2 + 10x3 + ............... [Ans. 3)x1(1 ]

(ii) Find the sum to n terms and also S(a) 1 + 5

4 + 257 + 35

10 + ........... [Ans. Sn = 1635 1n516

7n12

; S = 1635 ]

(b) 53 + 15

5 + 457 + 135

9 + ...... [Ans. Sn =

n32n25

3 ]

(iii) 193 + 219

33 + 319333 + 419

3333 + ..........

[Hint:

......10

11019

11019

11093

33

22

=

......19

101910

19109

3 32 ]

6. Miscellaneous sequences(Type-1): Sequence dealing with n ; 2n ; 3n

(1) 2)1n(nn

* 6)1n2)(1n(nn2 .rememberedbetoResultsdone.betoProof

** 223 n2)1n(nn

For proof :* Consider the identity k3 (k 1)3 = 3k2 3k + 1** Consider the identity k4 (k 1)4 = 4k3 6k2 + 4k 1Note : (i)

n

1r rr)ba( =

n

1r rn

1r rba ; (ii)

n

1r rak =

n

1r rak

(iii)

n

1rk =

n

1rk = k n


for e.g.

(1)

n

1i

i

1j

j

1k1 =

n

1i

i

1j)j( =

n

1i 2)1i(i = nn21 2

=

2)1n(n

6)1n2)(1n(n

21 = 12

)4n2)(1n(n = 6)2n)(1n(n

(2)117/QE Find the value(s) of the positive integer n for which the quadratic equation,n10)kx)(1kx(n

1k

has solutions and + 1 for some . [Ans. 11]

[Hint: )Kx()Kx(n1K

2

= 10 n

)1K(Kx)1K2(xn1K

2

= 10n

nx2 + n2x + 2)1n(n

6)1n2)(1n(n 10n = 0

3x2 + 3nx + (n2 31) = 0

331n2.

n)1(2

now proceed n = 11 or 11]1]

EXAMPLES :TYPE-1 (Base on 2n ; 3n ) :(i) Compute the sum

(31)2 + (32)2 + (33)2 + ....... + (50)2[Hint: (12 + 22 + ..... + (50)2 ) (12 + 22 + ..... + (30)2 )](ii) (a) Compute the sum of the series whose nth term is given by

Tn = n(n + 1)(3n 1)(b) 32 + 72 + 112 + ....... (sum to n terms)[Hint: Tn = (4n 1)2 ; Sn = ]

H.W. after 5th lecture : Complete Ex. 5(b) (Hall & Knight)6TH LECTURE

TYPE-2 (Using method of difference) :If T1, T2, T3, ....... are the terms of a sequence then the termsT2 T1, T3 T2, T4 T3 ..........some times are in A.P. and some times in G.P. For such series we first compute theirnth term and then compute the sum to n terms, using sigma notation.

EXAMPLES :(i) 6 + 13 + 22 + 33 + .................. n terms [Ans. n6

)13n6)(1n(n ]

(ii) 3 + 8 + 15 + 24 + ....... up to n terms [Ans. 6)7n2)(1n(n ]


(iii) 5 + 7 + 13 + 31 + 85 + ........... up to n terms. [Ans. 4n + 21 (3n + 1)]

(iv) 2 + 5 + 14 + 41 + 122 + ........ up to n terms. [Ans. 41 (3n + 1 + 2n 3)]

(v) 1 +

311 +

23

1311 + ........ +

1n2 3

1.....31

311 [Ans. 1n

n

341)1n2(3

]

TYPE -3 (Splitting the nth term as a difference of two) :(a) Here is a series in which each term is composed of the reciprocal of the product of r

factors in A.P., the first factor of the several terms being in the same A.P.EXAMPLES :(i) 4321

1 + 54321 + 6543

1 + ......[Ans. Sn =181 )3n)(2n)(1n(3

1 ; S=18

1 ]

(ii) 5311 + 753

1 + ........+ 9751 + .........

(iii) In case a factor is missing

e.g. 4213 + 532

4 + 6435 + ........ then split the nth as given below

[Hint :Tn = )3n)(1n(n2n

= )3n)(2n)(1n(n)2n( 2

(Note: Nr must be free from n and Dr must contain consecutive factors)

= )3n)(2n)(1n(n4n4n2

= )3n)(2n)(1n(n4n3)1n(n

= )2n)(1n(1

+ )3n)(2n)(1n(3

+ )3n)(2n)(1n(n4

(iv) ........975314

75313

5312

311

(v) Find Sn and S for ...........8642531

64231

421

[Hint: Tn = )2n2.........(642)]1n2()2n2)[(1n2......(531

= )2n2.........(642

)1n2......(531n2.........642)1n2......(531

]

TYPE-4 :Here is a series in which each terms is composed of r factor in A.P., the first factor ofthe several terms being in the same A.P.

Examples :(i) 1234 + 2345 + 3456 + ........ up to n terms[Sol. Tn = n(n + 1)(n + 2)(n +3)

= 5)]1n()4n)[(3n)(2n)(1n(n


= 51 [n(n + 1)(n + 2)(n + 3)(n + 4) (n 1)n(n +1)(n + 2)(n + 3)]

T1 = 51 [12345 0]

T2 = 51 [23456 12345]

Tn = 5

1 [n(n + 1)(n + 2)(n + 3)(n + 4) (n 1)n(n +1)(n + 2)(n + 3)]Sn = 5

1 [n(n + 1)(n + 2)(n + 3)(n + 4)] Ans. ]H.W. after 6th lecture : Ex.-29(a) complete and selected from Ex.-29(b) (Hall & Knight)

7TH LECTURE7(a) HARMONICAL PROGRESSION (HP) :

Definition : A sequence is said to be in H.P. if the reciprocals of its terms are in A.P.

e.g. if a1, a2, a3, ....... are in H.P. then 1a1 ,

2a1 ,

3a1 ..... are in A.P..

A standard H.P. is a1 + da

1 + d2a

1 + ........ + d1na

1

Note :(i) If the term of an H.P. is this means that the corresponding term of the A.P. is zero.(ii) There is no general formula for finding the sum to n terms of H.P.(iii) If a, b, c are in H.P. a

1 , b1 , c

1 are in A.PA.P

b2 = a

1 + c1 b = ca

ac2

or b1 a

1 = c1 b

1 i.e. abba = bc

cb i.e. ca = cb

ba

EXAMPLES :(i) If the 3rd , 6th and last term of a H.P. are 3

1 , 51 , 203

3 , find the number of terms.[Ans. 100]

(ii) If mth term of an H.P. is n, and nth term is equal to m then prove that (m + n)th termis nm

mn .

(iii) If a1, a2, a3......an are in H.P. then prove thata1a2 + a2a3 + a3a4 + ...... + an1an = (n 1)a1an[Hint:

1a1 ,

2a1 ,

3a1 ........

na1 are in A.P.A.P.

d =2121

aaaa

=3232

aaaa

= ...........1nnn1n

aaaa


now LHS = d1 [(a1 a2) + (a2 a3) + ...... + (an1 an)] = d

aa n1

nowna1 =

1a1 + (n 1)d (n 1)d =

1nn1

aaaa

d = )1n)(aa(aa

1nn1

; LHS = (n 1)a1an = RHS ]

7(b) HARMONICAL MEAN :If a, b, c are in H.P. then middle term is called the harmonic mean between them. Henceif H is the harmonic mean (H.M.) between a and b then a, H, b are in H.P.and H = ba

ab2 .

To insert n HM between a and b.Let H1, H2........Hn are n HM's between a and bhence a, H1, H2, .....Hn b are in H.P.

a1 ,

1H1 ,

2H1 ......

nH1 ,

b1 are in A.P.A.P.

b1 = a

1 + (n + 1)d ; a1 = (n + 1)d ; d = )1n(ab

ba

1H1 = a

1 + d

2H1 = a

1 + 2d

3H1 = a

1 + 3d

nH1 = a

1 + nd

n

1H1

i i = a

n + 2)1n)(n(d = a

n + 2)1n(n )1n(ab

ba

= n

ab2ba

a1 = ab2

n [2b + a b] = ab2)ba(n = n H

1

Hence sum of the reciprocals of all the n HM's between a and b is equal to n times asingle HM between a and b.

EXAMPLE :(i)130/1 If a is the A.M. of b and c ; b the G.M. of c and a , then prove that c is theH.M. of a and b.[Sol. 2a = b + c ....(1) ; b2 = ac ....(2)

T.P.T. c = baab2

Now 2ab = (b + c)b = b2 + bc = ac + bc = c (a+b) ]


(ii) If a2, b2, c2 are in A.P. show that b + c, c + a, a + b are in H.P.[Sol. By adding ab + ac + bc to each term, we see that

a2 + ab + ac + bc, b2 + ba + bc + ac, c2 + ca + cb + ab are in A.P.that is (a + b)(a + c), (b + c)(b + a), (c + a)(c + b) are in A.P. dividing each terms by (a + b)(b + c)(c + a)

cb1 , ac

1 , ba

1 are in A.P..

that is, b + c, c + a, a + b are in H.P. ]Relation between A.M, G.M. and H.M :If a and b are two positive numbers then A G H and A, G, H are in G.P. i.e. G2=AH

Proof: We have A = 2ba , G = ab and H = ba

ab2 H bAa

G

now AH = ab = G2 A G H are in G.P.also G

A = HG ; A G G H

Hence A G H Infact RMS AM GM HM ]GENERAL ILLUSTRATIONS :(i) If xp

xa = yqya = zr

za and p, q, r are in A.P. then prove that x, y, z are in H.P..

[Sol. p1x

a = q

1ya

= r1z

a ; qp

ya

xa

= rqza

ya

Since p, q, r are in A.P p q = q r x

a ya = y

a za x

1 , y1 , z

1 are in A.P.. x, y, z in H.P. ]

(ii) If ax = by = cz = dw = .... and a, b, c, d are in G.P. then prove that x, y, z, w... are in H.P.[T/S Ex-1]

[Sol. If ax = by = cz = dw = ....... = k

a = x1

k ; b = y1

k ; c = z1

k ; d = w1

k a, b, c, d are in G.P. a

b = bc = c

d

i.e. z1w1

y1z1

x1y1

kk

kk

kk = .........

z1

y1

k = y

1z1

k = z

1w1

k

y1 x

1 = z1 y

1 = w1 z

1

x1 , y

1 , z1 , w

1 , .... are in A.P.A.P. x, y, z, w ..... H.P.]


(iii) If a, b, c are three distinct positive reals in H.P. then prove that an + cn > 2bn.[Sol. using AM > GM in an and cn

2ca nn > (an cn)1/2

i.e. an + cn > 2(ac)n/2 ....(1)a, b, c are in H.P. b is the HM between a and c.

hence applying GM > HMac > b ; (ac)n/2 > bn ; 2(ac)n/2 = 2bn ....(2)

from (1) and (2)an + cn > 2bn ]

(iv) If a, b, c are in H.P. p, q, r are in H.P. and ap, bq, cr are in G.P. then prove that

rp + p

r = ca + a

c

[Sol. b = 2ca ; q = rp

pr2 and apcr = b2q2

apcr =22

rppr22

ca

= 2

222

)rp(rp)ca(

= pr)rp( 2 = ac

)ca( 2

= pr

rp = a

cca Hence proved. ]

H.W. after 7th lecture : Complete Ex.-6(a) (Hall & Knight) + Ex-1 complete8TH LECTURE : Ex-3 and Ex-2 if possible

sequence and series

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