Important notes on Sequence and Series

Important Concepts and Formulas – Sequence and Series Arithmetic Progression(AP)

 

Arithmetic progression(AP) or arithmetic sequence 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

Examples

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

If a, b, c are in AP, 2b = a + c

nth term of an arithmetic progression

tn = a + (n – 1)d

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

Number of terms of an arithmetic progression

n=[(la)/d]+1

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

n=[(la)/d]+1=(728)/4+1=64/4+1=16+1=17

Sum of first n terms in an arithmetic progression

Sn=n/2[ 2a+(n1)d ] =n/2(a+l)

where 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
=n/2[2a+(n1)d]=20/2[(2×4)+(201)3]=10[8+(19×3)]=10(8+57)=650
Example 2
Find 6 + 9 + 12 + . . . + 30

a = 6
l = 30
d = 9 – 6 = 3

n=(la)/d+1=(306)/3+1=243+1=8+1=9

Sum, S
=n/2(a+l)=9/2(6+30)=9/2×36=9×18=162

Arithmetic Mean

If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c. In this case, b=12(a+c)

The Arithmetic Mean (AM) between two numbers a and b = 1/2(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.

Additional Notes on APTo solve most of the problems related to AP, the terms can be conveniently 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 AP is increased, decreased , multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.


In an AP, sum of terms equidistant from beginning and end will be constant.

Harmonic Progression(HP)Non-zero numbers a1, a2, a3,  ana1, a2, a3, ⋯ an are in Harmonic Progression(HP) if 1a1, 1a2, 1a3, 1an1a1, 1a2, 1a3, ⋯1an are in AP. Harmonic Progression is also known as harmonic sequence.

Examples

12,16,110,12,16,110,⋯ is a harmonic progression (HP)

Three non-zero numbers a, b, c will be in HP, if  1a, 1b, 1c1a, 1b, 1c are in AP

If a, (a+d), (a+2d), . . . are in AP, nthterm of the AP = a + (n – 1)d

Hence, if 1a,1a+d,1a+2d,1a,1a+d,1a+2d,⋯ are in HP, nthterm of the HP = 1a+(n1)d1a+(n−1)d

If a, b, c are in HP, b is the Harmonic Mean(HM) between a and c

In this case, b=2ac/(a+c)

The Harmonic Mean(HM) between two numbers a and b = 2ab/(a+b)
If a, a1, a2 … an, b are in HP we can say that a1, a2 … an are the n Harmonic Means between a and b.
If a, b, c are in HP, 2/b=1/a+1/c

geometric progression(GP)Geometric Progression(GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant.

A geometric progression(GP) is given by a, ar, ar2, ar3, …
where a = the first term , r = the common ratio

Examples

1, 3, 9, 27, … is a geometric progression(GP) with a = 1 and r = 3

2, 4, 8, 16, … is a geometric progression(GP) with a = 2 and r = 2

If a, b, c are in GP, b2 = ac

nth term of a geometric progression(GP)

tn=arn-1

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 = 4/2 = 2,   n = 10

10th term, t10
=arn-1=2×2101=2×29=2×512=1024=arn−1=2×210−1=2×29=2×512=1024
Example 2
Find 5th term in the series 5, 15, 45, …

a = 5,   r = 155155 = 3,   n = 5

5th term, t5
=arn-1=5×35−1=5×34=5×81=405

Sum of first n terms in a geometric progression(GP)

Sn=[arn-1/r1 (if r>1)

             a(1rn)/1r (if r<1)]

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 = 12/4 = 3,   n = 6

Here r > 1. Hence,
S6=a(rn1)r1=4(361)31=4(7291)2=4×7282=2×728=1456S6=a(rn−1)r−1=4(36−1)3−1=4(729−1)2=4×7282=2×728=1456
Example 2
Find 1+1/2+1/4+ … up to 5 terms

a = 1,   r = (1/2)1=1/2,   n = 5

Here r < 1. Hence,
S6=a(1rn)/1r=1[1(1/2)^5]/(11/2)=(11/32)(1/2)=1(15/16)

Sum of an infinite geometric progression(GP)

S=a/1r  (if    -1 < r < 1)
where a= the first term , r = common ratio

Example
Find 1+1/2+1/4+1/8+

a = 1,       r = (1/2)1=1/2

Here -1 < r < 1. Hence,
S=a/1r=1(11/2)=1(1/2)=2

Geometric Mean

If three non-zero numbers a, b, c are in GP, b is the Geometric Mean(GM) between a and c. In this case, b=acb=ac

The Geometric Mean(GM) between two numbers a and b = abab

(Note that if a and b are of opposite sign, their GM is not defined.)

Additional Notes on GPTo solve most of the problems related to GP, the terms of the GP can be conveniently taken as
3 terms: arar, a, ar
5 terms: a/r2a/r, a, ar, ar2


If a, b, c are in GP,( ab)/(bc)=a/b


In a GP, product of terms equidistant from beginning and end will be constant.

Relationship Between Arithmetic Mean, Harmonic Mean, and Geometric Mean of Two NumbersIf GM, AM and HM are the Geometric Mean, Arithmetic Mean and Harmonic Mean of two positive numbers respectively, then

GM2 = AM × HM

Some Interesting Properties to Note

Three numbers a, b and c are in AP if b=(a+c)/2

Three non-zero numbers a, b and c are in HP if b=2ac/(a+c)

Three non-zero numbers a, b and c are in HP if ab/bc=a/c

Let A, G and H be the AM, GM and HM between two distinct positive numbers. Then

(1) A > G > H
(2) A, G and H are in GP

If a series is both an AP and GP, all terms of the series will be equal. In other words, it will be a constant sequence.
Power Series : Important formulas
1+1+1+ n terms=1=n1+2+3+⋯+n =n=n(n+1)/212+22+32+⋯+n2 =∑n2=n(n+1)(2n+1)/613+23+33+⋯+n3=∑n3=n2(n+1)/24=[n(n+1)/2]2

Find the nth term for the AP : 11, 17, 23, 29, …

Find the sum of the AP in the above question till first 10 terms.

Find the sum of the series 32, 16, 8, 4, … upto infinity.

The sum of three numbers in a GP is 26 and their product is 216.F ind the numbers.

1+2+3+.........+100

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