Progression of sequence and series
(scroll down for class work solved questions)
Arithmetic Progression (A.P.) (or Arithmetic Sequence)
- nth term of an arithmetic progression
- Number of terms of an arithmetic progression
- Sum of first n terms in an arithmetic progression
- Arithmetic Mean
- If a, b, c are in AP, 2b = a + c
- To solve most of the problems related to A.P., the terms can be
conveiently taken as
3 terms : (a – d), a, (a +d)
4 terms : (a – 3d), (a – d), (a + d), (a +3d)
5 terms : (a – 2d), (a – d), a, (a + d), (a +2d) - Tn = Sn - Sn-1
- If each term of an A.P. is increased, decreased , multiplied or
divided by the same non-zero constant, the resulting sequence also
will be in A.P. - In an A.P., sum of terms equidistant from beginning and end will be constant
Harmonic Progression (H.P.) (or Harmonic Sequence)
Geometric Progression (G.P.) (or Geometric Sequence)
- nth term of a geometric progression (G.P.)
- Sum of first n terms in a geometric progression (G.P.)
- Sum of an infinite geometric progression (G.P.)
- Geometric Mean
- If a, b, c are in G.P., b2 = ac
- If a, b, c are in G.P.,
a−bb−c=ab - In a G.P., product of terms equidistant from beginning and end will be constant.
- To solve most of the problems related to G.P., the terms of the G.P. can be conveiently taken as
3 terms :ar , a, ar
5 terms :ar2 ,ar , a, ar, ar2 Relationship Between Arithmetic Mean, Harmonic Mean,
and Geometric Mean of Two Numbers
Some Interesting Properties to Note
- A > G > H
- A, G and H are in GP
Power Series : Important formulas
Arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, d to the preceding term. The constant d is called common difference.
An arithmetic progression is given by a, (a + d), (a + 2d), (a + 3d), ...
where a = the first term , d = the common difference
where a = the first term , d = the common difference
Examples for Arithmetic Progressions
1, 3, 5, 7, ... is an arithmetic progression (AP) with a = 1 and d = 2
7, 13, 19, 25, ... is an arithmetic progression (AP) with a = 7 and d= 6
tn = a + (n – 1)d
where tn = nth term, a= the first term , d= common difference
where tn = nth term, a= the first term , d= common difference
Example 1 : Find 10th term in the series 1, 3, 5, 7, ...
a = 1
d = 3 – 1 = 2
10th term, t10 = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19
Example 2 : Find 16th term in the series 7, 13, 19, 25, ...
a = 7
d = 13 – 7 = 6
16th term, t16 = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97
where n = number of terms,
a= the first term ,
l = last term,
d= common difference
Example : Find the number of terms in the series 8, 12, 16, . . .72
a = 8
l = 72
d = 12 – 8 = 4
a = the first term,
d= common difference,
l=tn=nth term = a+(n−1)d
Example 1 : Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms
a = 4
d = 7 – 4 = 3
Sum of first 20 terms, S20 =n2[ 2a+(n−1)d ]=202[ (2×4)+(20−1)3 ]
Example 2 : Find 6 + 9 + 12 + . . . + 30
a = 6
l = 30
d = 9 – 6 = 3
If a, b, c are in AP, b is the Arithmetic Mean (A.M.) between a and c.
In this case,b=12(a+c)
In this case,
The Arithmetic Mean (A.M.) between two numbers a and b = 12(a+b)
If a, a1, a2 ... an, b are in AP we can say that a1, a2 ... an
are the n Arithmetic Means between a and b.
are the n Arithmetic Means between a and b.
Examples for Harmonic Progressions
Three non-zero numbers a, b, c will be in HP, if 1a, 1b, 1c are in A.P.
If a, (a+d), (a+2d), . . . are in A.P., nthterm of the A.P. = a + (n - 1)d
Hence, if1a,1a+d,1a+2d,⋯ are in H.P., nthterm of the H.P. = 1a+(n−1)d
Hence, if
If a, b, c are in HP, b is the Harmonic Mean(H.M.) between a and c
In this case,b=2aca+c
In this case,
The Harmonic Mean (H.M.) between two numbers a and b = 2aba+b
If a, a1, a2 ... an, b are in H.P. we can say that a1, a2 ... an are the n Harmonic Means between a and b.
If a, b, c are in HP, 2b=1a+1c
A sequence of non-zero numbers is a Geometric Progression (G.P.) if the ratio of any term and its preceding term is always constant.
A Geometric Progression (G.P.) is given by a, ar, ar2, ar3, ...
where a = the first term , r = the common ratio
where a = the first term , r = the common ratio
Examples for Geometric Progressions
1, 3, 9, 27, ... is a geometric progression (G.P.) with a = 1 and r = 3
2, 4, 8, 16, ... is a geometric progression (G.P.) with a = 2 and r = 2
where tn = nth term, a= the first term , r = common ratio, n = number of terms
Example 1 : Find the 10th term in the series 2, 4, 8, 16, ...
a = 2, r = 42 = 2, n = 10
10th term, t10 = arn−1=2×210−1=2×29=2×512=1024
Example 2 : Find 5th term in the series 5, 15, 45, ...
a = 5, r = 155 = 3, n = 5
5th term, t5 = arn−1=5×35−1 = 5 × 34 = 5 × 81 = 405
where a= the first term , r = common ratio, n = number of terms
Example 1 : Find 4 + 12 + 36 + ... up to 6 terms
a = 4, r = 124 = 3, n = 6
Here r > 1. Hence, S6=a(rn−1)r−1=4(36−1)3−1=4(729−1)2=4×7282=2×728=1456
Example 2 : Find 1+12+14+ ... up to 5 terms
a = 1, r = (12)1=12, n = 5
Here r < 1. Hence, S6=a(1−rn)1−r=1[1−(12)5](1−12)=(1−132)(12)=(3132)(12)=3116=11516
where a= the first term , r = common ratio
Example : Find 1+12+14+18+ . . . ∞
a = 1, r = (12)1=12
Here 0 < r < 1. Hence, S∞=a1−r=1(1−12)=1(12)=2
If three non-zero numbers a, b, c are in G.P., b is the Geometric Mean (G.M.) between a and c.
In this case,b=ac−−√
In this case,
The Geometric Mean (G.M.) between two numbers a and b = ab−−√
(Note that if a and b are of opposite sign, their G.M. is not defined.)
(Note that if a and b are of opposite sign, their G.M. is not defined.)
If GM, AM and HM are the Geometic Mean, Arithmetic Mean and Harmonic Mean of two positive numbers respectively, then
GM2 = AM × HM
GM2 = AM × HM
Three numbers a, b and c are in AP if b=a+c2
Three non-zero numbers a, b and c are in HP ifb=2aca+c
Three non-zero numbers a, b and c are in HP ifa−bb−c=ac
Three non-zero numbers a, b and c are in HP if
Three non-zero numbers a, b and c are in HP if
Let A, G and H be the A.M., G.M. and H.M. between two distinct positive numbers. Then
If a series is both an A.P. and G.P., all terms of the series will be equal. In other words, it will be a constant sequence.
1+1+1+⋯ n terms=∑1=n 1+2+3+⋯+n=∑n=n(n+1)2 12+22+32+⋯+n2=∑n2=n(n+1)(2n+1)6 - 13+23+33+⋯+n3=∑n3=n2(n+1)24=[n(n+1)2]2
class work