High Level Number series for IBPS Clerk

 Number series for IBPS Clerk 

 

  Number series for IBPS Clerk  

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 basics of 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 – 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.

  1. It might be a double or triple series. Here every alternate number or every 3rd number form a series
  2. 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.

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=(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 = 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]=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)

The Arithmetic Mean (AM) 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.

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, ⋯ 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)

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

The Harmonic Mean(HM) between two numbers a and b = 2aba+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, 2b=1a+1c

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.

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 = 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

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.)

Additional Notes on GPTo solve most of the problems related to GP, the terms of the GP can be conveniently taken as
3 terms: ar, a, ar
5 terms: ar2ar, a, ar, ar2


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 

GM2 = AM × HM

Some Interesting Properties to Note

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

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  

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