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Sequences and Series
Session MPTCP05
1. Revisit G.P. and sum of n terms of a G.P.
2. Sum of infinite terms of a G.P.3. Geometric Mean (G.M.) and
insertion of n G.M.s between two given numbers
4. Arithmetico-Geometric Progression (A.G.P.) - definition, nth term
5. Sum of n terms of an A.G.P.6. Sum of infinite terms of an A.G.P.7. Harmonic Progression (H.P.) -
definition, nth term
Session Objectives
Geometric Progression
A sequence is called a geometric progression (G.P.) if the ratio between any term and the previous term is constant.
_I005
The constant ratio, generally denoted by r is called the common ratio.
a1 = a
a2 = ar
a3 = ar2
a4 = ar3
an = ar(n-1)
First term
General Term
Problem Solving Tip
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
_I005
Important Properties of G.P.s
a, b, c are in G.P. b2 = ac _I005
Sum of n Terms of a G.P.
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)
n
n
r 1S a
r 1
_I006
Sum of Infinite Terms of a G.P.
Sum of n terms of a G.P.,
n n n
n
r 1 1 r a arS a a
r 1 1 r 1 r 1 r
Case(i) r 1nar
as n , 01 r
n
nn n
a ar aS lim S lim
1 r 1 r 1 rCase(ii) r 1
naras n ,
1 r
n
nn n
a arS lim S lim
1 r 1 r
_I007
aS , r 1
1 r
S , r 1
Illustration
_I007
Single Geometric Mean
G is the G.M. of a and b
G2 = ab
_I008
Geometric Mean – a Definition
If n terms G1, G2, G3, . . . Gn are inserted between two numbers a and b such that a, G1, G2, G3, . . . , Gn, b form a G.P.,
_I008
then G1, G2, G3, . . . , Gn are called geometric means (G.M.s) of a and b.
Geometric Mean – Common Ratio
Let n G.M.s be inserted between two numbers a and b
The G.P. thus formed will have (n+2) terms.
Let the common ratio be r
Now b = ar(n+2-1) = ar(n+1)
1 mn 1 n 1
mb b
r , G aa a
_I008
Property of G.M.s
Let n G.M.s G1, G2, G3, . . ., Gn be inserted between a and b.
Then,
n
n21 2 3 nG G G ...G G ab
_I008
Illustrative Problem
Q. Insert 3 G.M.s between 4 and 9
A. Let the required G.M.s be G1,
G2 and G3.
Common ratio r =
1
49 3
4 2
13
G 4 2 62
23
G 4 62
33 3
G 4 3 62 2
_I008
Illustrative Problem
Q. If the A.M. between a and b is twice as great as the G.M., a:b is equal to
(a) (b)
(c) (d)
2 3
2 3
7 4 3
7 4 3
2
7 4 3
2
3
_I008
Illustrative Problem
A. Given that
Squaring both sides, we get,
a b2 ab
2
a b 4 ab
2 2a 2ab b 16ab Dividing by b2 and putting =r, we get,
ab
2r 14r 1 0 14 196 4
r2
r 7 4 3
_I008Q. If the A.M. between a and b is twice as
great as the G.M., a:b is equal to
Illustrative Problem
Ans : (a)
A.22
22
2 3 2.2. 3r
2 3
22 3
r2 3 2 3
2 3 ar
b2 3
_I008Q. If the A.M. between a and b is twice as
great as the G.M., a:b is equal to
r 7 4 3
Arithmetico-Geometric Progression
A sequence is called an arithmetico-geometric progression (A.G.P.) if the nth term is a product of the nth term of an A.P. and the nth term of a G.P.
_I009
a1 = a
a2 = (a+d)r
a3 = (a+2d)r2
a4 = (a+3d)ar3
an = {a+(n-1)d}r(n-1)
First term
General Term
Sum of n Terms of an A.G.P.
Consider an A.G.P. with general
term {a+(n-1)d}r(n-1).
Let the sum of first n terms be Sn
n 12nS a a d r a 2d r ... a n 1 d r
2 n 1 nnrS ar a d r ... a n 2 d r a n 1 d r
2 n 1 nn nS rS a dr dr ... dr a n 1 d r
nn 1
n 2
a n 1 d ra 1 rS dr
1 r 1 r1 r
n 1n
n1 r
1 r S a dr a n 1 d r1 r
_I010
Illustrative Problem
Q. Find the sum of the first 10 terms of the given sequence :
1, 3x, 5x2, 7x3, . . .
A. Let
S = 1+3x+5x2+7x3+ . . . +{1+(10-1)2}x(10-1)
S = 1+3x+5x2+7x3+ . . . +19x9
xS = x+3x2+5x3+ . . . +17x9 +19x10
S-xS = 1+(2x+2x2+2x3+ . . . 2x9)-19x10
9
101 x
1 x S 1 2x 19x1 x
910
2
1 x1 19xS 2x
1 x 1 x
_I010
Sum of Infinite Terms of an A.G.P.
Sum of n terms of an A.G.P.,
Case(i) r 1
nn 1
2 2
a n 1 d r1 r dras n , dr , 0
1 r1 r 1 r
nn 1
n 2n n
a n 1 d ra 1 rS lim S lim dr
1 r 1 r1 r
n 1 n
n 2
a 1 r {a (n 1)d}rS dr
1 r 1 r1 r
2a d r
S , r 11 r 1 r
_I011
Sum of Infinite Terms of an A.G.P.
Case(ii) r 1
nn 1
2
a n 1 d r1 ras n , dr ,
1 r1 r
nn 1
n 2n n
a n 1 d ra 1 rS lim S lim dr
1 r 1 r1 r
S , r 1
Sum of n terms of an A.G.P.,
n 1 n
n 2
a 1 r {a (n 1)d}rS dr
1 r 1 r1 r
_I011
Illustrative Problem
Q. The sum to infinity of the series
is2 3
4 7 101 ...
5 5 5
(a) 16/35 (b) 11/8
(c) 35/16 (d) 8/6
_I011
Illustrative Problem
A. Let the required sum be S
Ans : (c)
2 3
4 7 10S 1 ...
5 5 5
2 3 4
S 1 4 7 10...
5 5 5 5 5
2 3
S 3 3 3S 1 ...
5 5 5 5
2
4S 3 1 1 3 5 71 1 ... 1 .
5 5 5 5 4 4535
S16
_I011Q. The sum to infinity of the series
is
2 3
4 7 101 ...
5 5 5
Illustrative Problem
Q. The sum of the infinite series 1 + (1+b)r + (1+b+b2)r2 + (1+b+b2+b3)r3 . . ., r and b being proper fractions is :
1
(a)1 r 1 br
1
(b)1 r 1 br
1
(c)1 r 1 br
1
(d)1 r 1 br
_I011
Illustrative Problem
A. Let the required sum be S
Subtracting, we have,
Ans : (a)
2 2 2 3 3S 1 1 b r 1 b b r 1 b b b r ...
2 2 3rS r 1 b r 1 b b r ...
2 2 3 3 4 41 r S 1 br b r b r b r ...
1
S1 r 1 br
_I011Q. The sum of the infinite series 1 + (1+b)r + (1+b+b2)r2 + (1+b+b2+b3)r3 . . . (r and b being proper fractions ) is :
Harmonic Progression
A sequence is called a harmonic progression (H.P.) if the reciprocals of its terms form an A.P.
_I012
11
aa
21
aa d
31
aa 2d
41
aa 3d
n1
aa n 1 d
First term
General Term
Class Exercise Q1.
Q. The first two terms of an infinite G.P. are together equal to 5, and every term is 3 times the sum of all the terms that follow it, the series is :
1 1(a) 4, 1, , ....
4 16
4 8(b) 3, 2, , , ....
3 9
(c) Either (a) or (b)
(d) None of these
_I007
Class Exercise Q1.
A. Let the first term of the G.P. be a and the common ratio be r.
Given that a+ar = 5 and
ar 1a 3 r
1 r 4
_I007Q. The first two terms of an infinite G.P. are
together equal to 5, and every term is 3 times the sum of all the terms that follow it, the series is :
Now,
aa 5 a 4
4
Ans : (a)
Class Exercise Q2.
_I007Q. Find the value of p, if S for the G.P.
2
1 1 25p,1, , , . . . is
p 4p
Class Exercise Q2.
_I007Q. Find the value of p, if S for the G.P.
21 1 25
p,1, , , . . . isp 4p
A. S for the given G.P.
p 1S common ratio
1 p1p
2p 25
p 1 4
24p 25p 25 0
p 5 4p 5 0
5
Ans. p 5, or p4
Class Exercise Q3.
_I008Q. If one G.M. G and two A.M.s p and q are inserted between two quantities, show that G2 = (2p-q)(2q-p).
Class Exercise Q3.
_I008Q. If one G.M. G and two A.M.s p and q are
inserted between two quantities, show that G2 = (2p-q)(2q-p).
A. Let the two quantities be a and b.
a, p, q, b are in A.P.b a
3
Common difference =
b a 2a bp a
3 3
b a a 2bq a 2
3 3
2
R.H.S 2p q 2q p
4a 2b a 2b 2a 4b 2a b
3 3
ab
G
L.H.S.
Q.E.D.
Class Exercise Q4.
_I008Q. n G.M.s are inserted between 16/27 and 243/16. If the ratio of the (n-1)th G.M. to the 4th G.M. is 9 : 4, find n.
Class Exercise Q4.
_I008Q. n G.M.s are inserted between 16/27 and
243/16. If the ratio of the (n-1)th G.M. to the 4th G.M. is 9 : 4, find n.
A. Common ratio1 1 8
n 1 n 1 n 1b 243 27 3
a 16 16 2
Given that 8 n 1 32
2n 1 n 13 2 9 3
.2 3 4 2
8 n 5
2n 13 3
2 2
n 5 1n 7
n 1 4
Class Exercise Q5.
_I010Q. Find the sum of the series :
3 5 71 . . . n terms
2 4 8
Class Exercise Q5.
_I010Q. Find the sum of the series :
3 5 71 . . . n terms
2 4 8
A. We see that
n n 1
2n 1a
2
n n 1
3 5 7 2n 1S 1 . . .
2 4 8 2
nn 1 n
2n 1 2n 1S 1 3 5. . .
2 2 4 8 2 2
n n 1 n
3 2 2 2 2 2n 1S 1 . . .
2 2 4 8 2 2
Class Exercise Q5.
_I010Q. Find the sum of the series :
3 5 71 . . . n terms
2 4 8
n n 1 n
3 2 2 2 2 2n 1S 1 . . .
2 2 4 8 2 2
n 1
n n
11
3 1 2n 12S 1 2.
12 2 212
n 1
n n3 2 1 2n 1S 1 1
2 3 2 2
n n 1
3 1 1 2 2n 1S
2 3 3 22
n n 12 1 1 6n
S9 9 2
Class Exercise Q6.
_I010Q. Find sum to n terms of the series :
1+2x+3x2+4x3+ . . . (x 1)
Class Exercise Q6.
_I010Q. Find sum to n terms of the series :
1+2x+3x2+4x3+ . . . (x 1)
A. We see that an = nxn-1
Sn = 1+2x+3x2+4x3+ . . . +nxn-1
xSn = x+2x2+3x3+. . . +(n-1)xn-1+nxn
(1-x)Sn = 1+(x+x2+x3+ . . . xn-1)-nxn
n 1n
n
1 x1 x S 1 x nx
1 x
n n 1
n 2
1 nx 1 xS x
1 x 1 x
Class Exercise Q7.
_I011Q. Find the sum of infinite terms of the series :
2 3
4 6 82 . . .
11 11 11
Class Exercise Q7.
_I011A.
2 3
4 6 8S 2 . . .
11 11 11
2 3
S 2 4 6. . .
11 11 11 11
Q. Find the sum of infinite terms of the series :
2 34 6 8
2 . . .11 11 11
2 3
10S 1 1 12 2 . . .
11 11 11 11
10S 2 112
11 11 10 121
S50
Class Exercise Q8.
_I011Q. Find the sum of the series :
2 3
3 5 71 . . .
2 2 2
Class Exercise Q8.
_I011Q. Find the sum of the series :
2 3
3 5 71 . . .
2 2 2
A. 2 3
3 5 7S 1 . . .
2 2 2
2 3
S 1 3 5. . .
2 2 2 2
2 3
3 2 2 2S 1 . . .
2 2 2 2
3 1 2 1S 1 1
12 3 31
2
2
S9
Class Exercise Q9.
_I012Q. If the pth, qth and rth terms of an H.P. be a, b, c respectively, then
(q-r)bc+(r-p)ca+(p-q)ab is equal to
(a) 1 (b) -1
(c) 0 (d) None of these
Class Exercise Q9.
_I012Q. If the pth, qth and rth terms of an H.P. be a,
b, c respectively, then
(q-r)bc+(r-p)ca+(p-q)ab is equal to
(a) 1 (b) -1
(c) 0 (d) None of theseA. The reciprocals the terms of the H.P. will be in A.P. Let this
A.P. have first term and common difference .
Given that
1
a ...(i)p 1
1
b ...(ii)q 1
1
c ...(iii)r 1
Class Exercise Q9.
_I012Q. If the pth, qth and rth terms of an H.P. be a,
b, c respectively, then
(q-r)bc+(r-p)ca+(p-q)ab is equal to
(a) 1 (b) -1
(c) 0 (d) None of theseA. Taking reciprocal of (i), (ii) and (iii), we have
1p 1 ...(iv)
a
1q 1 ...(v)
b
1r 1 ...(vi)
c
Class Exercise Q9
_I012Q. If the pth, qth and rth terms of an H.P. be a,
b, c respectively, then
(q-r)bc+(r-p)ca+(p-q)ab is equal to
(a) 1 (b) -1
(c) 0 (d) None of theseA. (iv)-(v), (v)-(vi), (vi)-(iv) gives,
b ap q ...(vii)
ab
c bq r ...(viii)
bc
a cr p ...(ix)
ac
Class Exercise Q9
_I012Q. If the pth, qth and rth terms of an H.P. be a,
b, c respectively, then
(q-r)bc+(r-p)ca+(p-q)ab is equal to
(a) 1 (b) -1
(c) 0 (d) None of theseA. (vii)c, (viii)a and (ix)b gives,
bc cap q ab
ca abq r bc
ab bcr p ac
Adding,
ca ab ab bc bc caq r bc r p ca p q ab
Class Exercise Q9
_I012Q. If the pth, qth and rth terms of an H.P. be a,
b, c respectively, then
(q-r)bc+(r-p)ca+(p-q)ab is equal to
(a) 1 (b) -1
(c) 0 (d) None of these ca ab ab bc bc caq r bc r p ca p q ab
A. (q-r)bc+(r-p)ca+(p-q)ab = 0
Ans : (c)
Class Exercise Q10.
Q. If ax = by = cz and x, y, z are in H.P. then a, b, c are in
(a) A.P. (b) H.P.
(c) G.P. (d) None of these
_I012
Class Exercise Q10.
A. ax = by = cz = (say)
Ans : (c)
11 1yx za , b , c
x, y, z are in H.P.1 1 1y 2x 2z
11 1 1 1 2
2x 2z x zb .
b ac
_I012Q. If ax = by = cz and x, y, z are in H.P. then
a, b, c are in
(a) A.P. (b) H.P.
(c) G.P. (d) None of these