## Number series for IBPS Clerk 2021

**Number series** questions can’t be avoided by the aspirant because these questions are asked every year. Number series questions are the questions that check your analyzing ability. **IBPS Clerk** Exam doesn’t just check only your accuracy but also check your speed.

In our previous posts, we have discussed the basics of the **Number series** and tips and tricks to solve number series questions. In this post, we will discuss a few **number series** questions that are frequently asked in **IBPS Clerk Exams**

## Different Type of Number Series

A series can be created in numerous ways. An understanding of these various ways can help us in recognizing the pattern followed in the number series. So here we go with some standard series types-

**Arithmetic Series** – The difference between successive terms is fixed. Subsequent terms are obtained by either adding or subtracting a fixed number. For example,

2, 5, 8, 11, 14, 17,….. Common Difference = 3

32, 25, 18, 11, 4,……. Common Difference = -7

**Geometric Series** – Each term of the series is obtained by multiplying (or dividing) the previous number by a fixed number. Hence the ratio between any 2 consecutive terms is the same. For example,

3, 6, 12, 24, 48, 96……. Common Ratio = 2

2048, 512, 128, 32……. Common Ratio = 1/4

**Arithmetic-Geometric Series** – Each term is first added (or subtracted) by a fixed number and then multiplied (or divided) by another number to obtain the subsequent term. For example,

4, 18, 60, 186….. => 4, (4+2)x3, (18+2)x3, (60+2)x3

**Geometric-Arithmetic Series** – Each term is first multiplied (or divided) by a fixed number and then added (or subtracted) by another number to obtain the subsequent term. For example,

3, 10, 24, 52…… => 3, (3×2)+4, (10×2)+4, (24×2)+4,…..

**Series of Squares, Cubes, etc.** – Each term is square or cube or a higher power of the previous term. For example,

3, 9, 81, 6561…. Each term is obtained by squaring the previous number

2, 8, 512, ……… Each term is obtained by cubing the previous number

Some non-standard ways in which series can be created –

**Series with subsequent Differences being in Arithmetic Progression (AP)** –

3, 7, 13, 21, 31, 43…… The differences in subsequent terms are 4, 6, 8, 10, 12…. which are in AP

**Series with Differences in Differences being in AP** –

336, 210, 120, 60, 24, 6, 0,…. The difference is 126, 90, 60, 36, 18, 6

The differences between differences being 36, 30, 24, 18, 12,…. and so on, which are in AP

**Inter-Mingled Series** – In this case, any two of the above series are mixed in one. For example,

1, 3, 5, 1, 9, -3, 13, -11, 17,….

Odd terms (1, 5, 9, 13, 17,….) of the series are in AP, whereas even terms (3, 1, -3, -11,…) are in geometric-arithmetic series in which subsequent terms are obtained by multiplying the previous term by 2 and then subtracting 5.

**How to solve Number series problems easily By Tricks**

13.2.15

Generally, two kinds of series are asked in the examination. One is based on numbers and the other based on alphabets.

Step 1: Observer is there any familiar numbers in the given series. Familiar numbers are primes numbers, perfect squares, cubes … which are easy to identify.

Step 2: Calculate the differences between the numbers. Observe the pattern in the differences. If the differences are growing rapidly it might be a square series, cube series, or multiplicative series. If the numbers are growing slowly it is an addition or subtraction series.

- It might be a double or triple series. Here every alternate number or every 3rd number form a series
- It might be a sum or average series. Here sum of two consecutive numbers gives a 3rd number. or an average of first two numbers give the next number

Step 3: Sometimes number will be multiplied and will be added another number So we need to check those patterns

#### Number Series Formula

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=644+1=16+1=17

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

Sn=n2[ 2a+(n−1)d ] =n2(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}

=n2[2a+(n−1)d]=202[(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

=n2(a+l)=92(6+30)=92×36=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, ⋯ an are in Harmonic Progression(HP) if 1a1, 1a2, 1a3, ⋯1an are in AP. Harmonic Progression is also known as harmonic sequence.

**Examples**

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

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

In this case, b=2aca+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 a 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=arn−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 = 42 = 2, n = 10

10^{th} term, t_{10}

=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 = 155 = 3, n = 5

5^{th} term, t_{5}

=arn−1=5×35−1=5×34=5×81=405

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

Sn={a(rn−1)r−1 (if r>1)a(1−rn)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 = 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

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

S∞=a1−r (if -1 < r < 1)

where a= the first term , r = common ratio

**Example**

Find 1+12+14+18+⋯∞

a = 1, r = (12)1=12

Here -1 < r < 1. Hence,

S∞=a1−r=1(1−12)=1(12)=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=ac

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

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

3 terms: ar, a, ar

5 terms: ar2, ar, a, ar, ar

^{2}

If a, b, c are in GP, a−bb−c=ab

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

Three numbers a, b and c are in AP if b=a+c2

Three non-zero numbers a, b and c are in HP if b=2aca+c

Three non-zero numbers a, b and c are in HP if a−bb−c=ac

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)212+22+32+⋯+n2 =∑n2=n(n+1)(2n+1)613+23+33+⋯+n3

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