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.

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

**n ^{th} term of an arithmetic progression**

t_{n} = a + (n – 1)d

where t_{n} = n^{th} term, a= the first term , d= common difference

**Example 1**

Find 10^{th} term in the series 1, 3, 5, 7, …

a = 1

d = 3 – 1 = 2

10^{th} term, t_{10} = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19

**Example 2**

Find 16^{th} term in the series 7, 13, 19, 25, …

a = 7

d = 13 – 7 = 6

16^{th} term, t_{16} = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97

**Number of terms of an arithmetic progression**

n=[(l−a)/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=[(l−a)/d]+1=(72−8)/4+1=64/4+1=16+1=17

**Sum of first n terms in an arithmetic progression**

Sn=n/2[ 2a+(n−1)d ] =n/2(a+l)

where a = the first term,

d= common difference,

l = t_{n} = n^{th} 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, S_{20}

=n/2[2a+(n−1)d]=20/2[(2×4)+(20−1)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=(l−a)/d+1=(30−6)/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)

If a, a_{1}, a_{2} … a_{n}, b are in AP we can say that a_{1}, a_{2} … a_{n} are the n Arithmetic Means between a and b.

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)

T_{n}= S_{n}– S_{n-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)

If a, (a+d), (a+2d), . . . are in AP, n^{th}term of the AP = a + (n – 1)d

Hence, if 1a,1a+d,1a+2d,⋯1a,1a+d,1a+2d,⋯ are in HP, n^{th}term of the HP = 1a+(n−1)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)

_{1}, a

_{2}… a

_{n}, b are in HP we can say that a

_{1}, a

_{2}… a

_{n}are the n Harmonic Means between a and b.

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.

^{2}, ar

^{3}, …

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

^{2}= ac

**n ^{th} term of a geometric progression(GP)**

tn=ar^{n-1}

where t_{n} = n^{th} term, a= the first term , r = common ratio, n = number of terms

**Example 1**

Find the 10^{th} term in the series 2, 4, 8, 16, …

a = 2, r = 4/2 = 2, n = 10

10^{th} term, t_{10}

=ar^{n-1}=2×210−1=2×29=2×512=1024=arn−1=2×210−1=2×29=2×512=1024

**Example 2**

Find 5^{th} term in the series 5, 15, 45, …

a = 5, r = 155155 = 3, n = 5

5^{th} term, t_{5}

=ar^{n-1}=5×35−1=5×34=5×81=405

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

^{n-1}/r−1 (if r>1)

a(1−r^{n})/1−r (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(rn−1)r−1=4(36−1)3−1=4(729−1)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(1−r^{n})/1−r=1[1−(1/2)^5]/(1−1/2)=(1−1/32)(1/2)=1(15/16)

**Sum of an infinite geometric progression(GP)**

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/1−r=1(1−1/2)=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.)

3 terms: arar, a, ar

5 terms: a/r

^{2}, a/r, a, ar, ar

^{2}

If a, b, c are in GP,( a−b)/(b−c)=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

GM^{2} = 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 a−b/b−c=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

1+1+1+⋯ n terms=∑1=n1+2+3+⋯+n =∑n=n(n+1)/21

^{2}+2

^{2}+3

^{2}+⋯+n

^{2}=∑n

^{2}=n(n+1)(2n+1)/61

^{3}+2

^{3}+3

^{3}+⋯+n

^{3}=∑n

^{3}=n

^{2}(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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