sequences and series session mptcp04. 1.finite and infinite sequences 2.arithmetic progression...
TRANSCRIPT
Sequences and Series
Session MPTCP04
1. Finite and infinite sequences2. Arithmetic Progression (A.P.) -
definition, nth term3. Sum of n terms of an A.P.4. Arithmetic Mean (A.M.) and
insertion of n A.M.s between two given numbers.
5. Geometric Progression (G.P.) - definition, nth term
6. Sum of n terms of a G.P.
Session Objectives
Sequence – a Definition
A sequence is a function whose domain is the set N of natural numbers.
_I001
a1, a2, a3, . . ., an, . . .
Finite and Infinite Sequences
Finite sequence :
a a1, a2, a3, . . ., an
Infinite sequence :
a a1, a2, a3, . . ., an, . . .
_I001
Series – a Definition
If
a1, a2, a3, . . ., an, . . .
is a sequence,
_I001
the expression
a1+a2+a3+ . . . +an+ . . .
is called a series.
Arithmetic Progression
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.
_I002
a1 = a
a2 = a+d
a3 = a+d+d = a+2d
a4 = a+d+d+d = a+3d
First termGeneral Term
an = a+d+d+d+... = a+(n-1)d
Is a Given Sequence an A.P.?
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.
_I002
Problem Solving Tip
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
_I002
Illustrative problem
Q. If sum of three numbers in A.P. is 45, and the second number is thrice the first number, find the three numbers.
_I002
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
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.
_I002
Important Properties of A.P.s
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
_I002
Important Properties of A.P.s
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
_I002
Important Properties of A.P.s
Example :
Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
a3+a8 = 22
_I002ak+an-(k-1) = a1+an
k = 2, 3, 4, . . . (n-1)
Important Properties of A.P.S
Example :
Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
a3+a8 = 22 = a5+a6 = 22
_I002ak+an-(k-1) = a1+an
k = 2, 3, 4, . . . (n-1)
Important Properties of A.P.s
Example :
Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
a3+a8 = 22 = a5+a6 = 22 = a1+a10 = 22
_I002ak+an-(k-1) = a1+an
k = 2, 3, 4, . . . (n-1)
Important Properties of A.P.s
a, b, c are in A.P. 2b = a+c _I002
Important Properties of A.P.s
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
_I002
Important Properties of A.P.s
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.
_I002
Sum of n Terms of an A.P.
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}
n 1n
S 2a n 1 d2
_I003
Sum of n Terms of an A.P.
n 1n
S 2a n 1 d2
This can also be written as :
n 1 1n
S a a n 1 d2
n 1 nn
S a a2
nn
S First Term Last Term2
_I003
Property of Sum of n Terms of an A.P.
A sequence is an A.P.
Sn = An2+Bn,
where A, B are constants.
2A is the common difference.
We know that, nn
S 2a n 1 d2
2n
d dS n a n
2 2Rearranging,
Or, Sn = An2+Bn.
_I003
Arithmetic Mean
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
_I004
a bA
2
Arithmetic Mean – a Definition
then A1, A2, A3, . . . , An are called arithmetic means (A.M.s) of a and b.
-4 -2 2 4A1 A2 A3 A4 A5
0-6
a
6
b
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.,
_I004
Arithmetic Mean – Common Difference
Let n A.M.s be inserted between two numbers a and b
Let the common difference be d
Now b = a+(n+2-1)d = a+(n+1)d
-4 -2 2 4A1 A2 A3 A4 A5
0-6
a
6
b
The A.P. thus formed will have (n+2) terms.
_I004
mb a b a
d ; A a mn 1 n 1
Property of A.M.s
Let n A.M.s A1, A2, A3, . . ., An be inserted between a and b.
Then,
1 2 3 na b
A A A ... A nA n2
_I004
Illustrative Problem
Q. Insert 3 A.M.s between -4 and 3
A. Let the required A.M.s be A1,
A2 and A3.
Common difference d = 3 4 7
4 4
17 9
A 44 4
214 1
A 44 2
321 5
A 44 4
_I004
Geometric Progression
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.)
_I005
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
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.
_I005
Important Properties of G.P.s
If G a, ar, ar2, . . ., arn-1
taking reciprocal of each term,
G’ is also a G.P. with a reciprocal common ratio.
2 n 1
1 1 1 1G' , , , . . .,
a ar ar ar
_I005
Important Properties of G.P.s
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.
_I005
Important Properties of G.P.s
akan-(k-1) = a1an
k = 2, 3, 4, . . . (n-1)
Example :
Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512
_I005
Important Properties of G.P.s
Example :
Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512
a3a8 = 512
_I005akan-(k-1) = a1an
k = 2, 3, 4, . . . (n-1)
Important Properties of G.P.s
Example :
Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512
a3a8 = 512 = a5a6 = 512
_I005akan-(k-1) = a1an
k = 2, 3, 4, . . . (n-1)
Important Properties of G.P.s
Example :
Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512
a3a8 = 512 = a5a6 = 512 = a1a10 = 512
_I005akan-(k-1) = a1an
k = 2, 3, 4, . . . (n-1)
Important Properties of G.P.s
a, b, c are in G.P. b2 = ac _I005
Important Properties of G.P.s
If G a, ar, ar2, . . ., arn-1
take every third term,
G’ a, ar3, ar6, . . .
G’ is also a G.P.
_I005
Important Properties of G.P.s
a1, a2, a3, . . . , an is a G.P. of positive terms
loga1, loga2, loga3, . . . logan is an A.P.
_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
Class Exercise Q1.
Q. If log 2, log (2x-1) and log (2x+3) are in A.P., find x.
_I002
Class Exercise Q1.Q. If log 2, log (2x-1) and log (2x+3) are in
A.P., find x. _I002A. Given that
log(2x-1)-log2 = log(2x+3)-log(2x-1)
x x
x
2 1 2 3log log
2 2 1
2x x 1 x 12 2 1 2 6 2x x2 4.2 5 0
x x2 5 2 1 0
x x2 5 2 cannot be negative
2log5
x log 5log2
Class Exercise Q2.
Q. Show that there is no infinite A.P. which consists only of distinct primes.
_I002
Class Exercise Q2.
_I002A. Let, if possible, there be an A.P.
consisting only of distinct primes : a1, a2, a3, . . ., an, . . .
an = a1+(n-1)d
Q. Show that there is no infinite A.P. which consists only of distinct primes.
1a 1 1 1a a a 1 1 d
1a 1 1a a (1 d)
Thus, (a1+1)th term is a multiple of a1.
Thus, no such A.P. is possible.
Q.E.D.
Class Exercise Q3.
_I003Q. , 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)
nn
S nP n 1 Q2
Class Exercise Q3.
_I003Q. , 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)
nn
S nP n 1 Q2
A. an = Sn - Sn-1
nn 1n
a nP n 1 Q n 1 P n 2 Q2 2
na P n 1 Q
n n 1d a a
d P n 1 Q P n 2 Q d Q Ans : (d).
Class Exercise Q4.
_I003Q. 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.
_I003Q. 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
Class Exercise Q5.
_I004Q. Find the value of n so that
be an A.M. between a and b (a, b are positive).
n 1 n 1
n n
a b
a b
Class Exercise Q5.
A. Given that,
_I004Q. Find the value of n so that
be an A.M. between a and b (a, b are positive).
n 1 n 1
n n
a b
a b
n 1 n 1
n n
a b a b
2a b
n 1 n 1 n na b a b ab
Dividing throughout by bn+1, we get,n 1 n
a a a1
b b b
na a a
1 1b b b
Class Exercise Q5.
_I004Q. Find the value of n so that
be an A.M. between a and b (a, b are positive).
n 1 n 1
n n
a b
a b
na a a
1 a b 1 1 0b b b
na a a1 1
b b b
n = 0
Class Exercise Q6.
_I004Q. 53 A.M.s are inserted between 2 and 98. Find the 27th A.M.
Class Exercise Q6.
_I004Q. 53 A.M.s are inserted between 2 and 98.
Find the 27th A.M.
A. Common differenceb a 98 2 48
n 1 53 1 27
2748
A a 27d 2 27 5027
Class Exercise Q7.
_I005Q. If the 3rd term of a G.P. is 4, what is the product of the first five terms?
Class Exercise Q7.
_I005Q. If the 3rd term of a G.P. is 4, what is the
product of the first five terms?
A. Let the first 5 terms of the G.P. be :
22
a a, , a, ar, arrr
Required product = a5
= (a3)5
=45
Class Exercise Q8.
_I005Q. If the 4th, 7th, 10th term of a G.P. are p, q, r respectively, then
(a) p2 = q2+r2 (b) q2 = pr
(c) p2 = qr (d) pqr+pq+q = 0
(M.N.R. 1995)
Class Exercise Q8.
_I005Q. If the 4th, 7th, 10th term of a G.P. are p, q, r
respectively, then
(a) p2 = q2+r2 (b) q2 = pr
(c) p2 = qr (d) pqr+pq+q = 0
(M.N.R. 1995)
A. Let the first term of the G.P. be and common ratio be .
Ans : (b)
p = 3, q = 6, r = 9
Now, pr = 212
= (6)2
= q2
Class Exercise Q9.
_I006Q. Find the sum to n terms of the sequence 6, 66, 666, . . .
Class Exercise Q9.
_I006Q. Find the sum to n terms of the sequence
6, 66, 666, . . .
A. Sn = (6+66+666+ . . .[n terms])
n6
S 9 99 999 . . . n terms9
n6
S 10 1 100 1 1000 1 . . . n terms9
2 3n
6S 10 10 10 . . . n terms n
9
n
n
10 16S 10 n
9 10 1
nn
6 10S 10 1 n
9 9
Class Exercise Q10.
_I006Q. How many terms of the G.P.
are needed to give the sum ?
2 1 1 3, , , , . . .
9 3 2 4
55
72
Class Exercise Q10.
_I006Q. How many terms of the G.P.
are needed to give the sum ?
2 1 1 3, , , , . . .
9 3 2 4
55
72
A. Common ratio1
332 29
Let the required number of terms be n.
n
n
n
31
55 2 2 2 32S 1
372 9 9 5 212
Class Exercise Q10.
_I006Q. How many terms of the G.P.
are needed to give the sum ?
2 1 1 3, , , , . . .
9 3 2 4
55
72n55 2 2 3
172 9 5 2
n
3 55 9 5 2431
2 72 2 2 32
n 53 3
2 2
n = 5
Thank you