# Tips And Tricks And Shortcut For Arithmetic Progressions

## Tips and Tricks for Arithmetic Progression in Aptitude

Arithmetic Progression can applied in real life by analyzing a certain pattern that we see in our daily life. It is also asked in Many Examination that’s why we need to know Tips and Tricks for Arithmetic Progression . If the initial term of an arithmetic progression is a, and the common difference of successive members is d, then the n-th term of the sequence is given by $a_{n}$ = a + (n − 1)d, n = 1, 2,..and so on

In this Page Tips and Tricks for Arithmetic Progression as well as Different Types of  Questions and their answers is given for Solving Questions of Arithmetic Progression in less Time.

### Tips and Tricks for Arithmetic Progression To Solve Arithmetic Progression in Aptitude

• An Arithmetic Progression or Arithmetic Sequence is a sequence of numbers or terms  in a way that the difference between the consecutive terms is constant. Here, tips and tricks for you on Arithmetic Progression questions quickly, easily, and efficiently in competitive exams.

An Arithmetic Progression is Represented in the form a, (a + d), (a + 2d), (a + 3d), …
where a = the first term, and d = the common difference.

n is number of Terms

General form, Tn = a + (n-1)d

Where Tn is nth term of an Arithmetic Progression

• There are 4 types of questions asked in exams.

### Type 1: Find nth term of series $t_{n} = a+(n-1)d$

Question 1 : Find 10th term in the series 1, 3, 5, 7, …

Options
A 20
B 19
C 15
D 21

Solution:    We know that,

tn = a + (n – 1)d
where tn = nth term,

a= the first term ,

d= common difference,

n = number of terms in the sequence

In the given series,

a (first term) = 1

d (common difference) = 2 (3 – 1, 5 – 3)

Therefore, 10th term = t10 = a + (n-1) d

t10 = 1 + (10 – 1) 2

t10 = 1 + 18

t10 = 19

Correct option: B

Question 2 : Find last term in the series if there are 8 term in this series 13 , 17 , 21 ,25….

Options
A 33
B 41
C 37
D 39

Solution:    We know that,

tn = a + (n – 1)d
where tn = nth term,

a = the first term ,

d = common difference,

n = number of terms in the sequence

In the given series,

a (first term) = 13

d (common difference) = 4(17 – 13, 21 – 17)

Therefore, 8th term = t8 = a + (n-1) d

t8 = 13 + (8 – 1) 4

t8 = 13 + 28

t8 = 41

Correct option: B

### Type 2: Find number of terms in the series $n = \frac{l-a}{d} +1$

Question 1 :  Find the number of terms in the series 7, 11, 15, . . .71

Options
A  12
B  25
C  22
D  17

Solution:    We know that, $n= [ \frac{(l-a)}{d} ]+ 1$

where n = number of terms,

a= the first term,

l = last term,

d= common difference

In the given series,

a (first term) = 7

l (last term) = 71

d (common difference) = 11 – 7 = 4

n=  [$\frac{(71-7)}{4}]+ 1$

n =  $\frac{64}{4} + 1$

n = 16 + 1

n = 17

Correct option: D

Question 2 :  Find the number of terms if First term = 22 ,Last term = 50 and common difference is 4

Options:

A  10
B  9
C  8
D  7

Solution:    We know that,

$n= [ \frac{(l-a)}{d} ]+ 1$

where n = number of terms,

a = the first term,

l = last term,

d = common difference

In the given series,

a (first term) = 22

l (last term) = 50

d (common difference) = 4

n = $[\frac{(50-22)}{4}]+ 1$

n = $\frac{28}{4} +1$

n = 7+ 1

n = 8

Correct option: C

### Type 3: Find sum of first ‘n’ terms of the series $S_{n} =\frac{n}{2}[2a+(n-1)\times d]$ or $\frac{n}{2}(a+l)$

Question 1. Find the sum of the series 1, 3, 5, 7…. 201

Options
A  12101
B  25201
C  22101
D  10201

Solution:    We know that,

Sn = $\frac{n}{2}[2a + (n − 1) d]$

OR

$\frac{n}{2} (a+l)$
where,

a = the first term,

d= common difference,

l = tn = nth term = a + (n-1)d

In the given series,

a = 1, d = 2, and l = 201

Since we know that, l = a + (n – 1) d

201 = 1 + (n – 1) 2

201 = 1 + 2n -2

202 = 2n

n = 101

Sn $\frac{ n}{2}(a+l)$

Sn =   $\frac{101}{2}$ (1+201)

Sn = 50.5 (1 + 201)

Sn = 50.5 x 202

Sn = 10201

Correct option: D

Question 2. Find the sum of the Arithmetic series if First term of this series is 45 , common difference is 5 and number of terms in this series is 8.                    Options
A  500
B  300
C  400
D  200

Solution:    We know that,

Sn = $\frac{n}{2}[2a + (n − 1) d]$

OR

$\frac{n}{2} (a+l)$
where, a = the first term,

d = common difference,

l = tn = nth term = a + (n-1)d

In the given series,

a = 45, d = 5, and n = 8

S$\frac{n}{2}[2a + (n − 1) d]$

Sn$\frac{8}{2}[2\times 45 + (8 − 1) \times 5]$

Sn = 4 (90 + 35)

Sn = 4 x 125

Sn = 500

Correct option: A

### Type 4: Find the arithmetic mean of the series. $b = \frac{a+c}{2}$

Question 1.Find the arithmetic mean of first five prime numbers.

Options:

A  6.6
B  3.6
C  5.6
D  7.6

Solution:    We know that

b$\frac{1}{2} (a + c)$

Here, five prime numbers are 2, 3, 5, 7 and 11

Therefore, their arithmetic mean (AM) = $\frac{(2+3+5+7+11)}{5} = 5.6$

Correct option: C

Question 2.Find Second Number if arithmetic mean of two numbers is 24 and First number is 10.

Options:

A  32
B  38
C  34
D 36

Solution:    We know that

b$\frac{1}{2} (a + c)$

Let second Number is b

Therefore, their arithmetic mean (AM) = $\frac{(10 + b)}{2} = 24$

b = 38

Correct option: B