# How To Solve Arithmetic Progression Questions Quickly

## How to solve Arithmetic Progressions questions easily

### Definition

Arithmetic progression (AP) 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 AP is represented in the form a, (a + d), (a + 2d), (a + 3d), …
where a = the first term, and  d = the common difference.
General form, Tn = a+ (n-1)d ## Type 1: Find nth term of an AP: tn = a + (n – 1)d

### Question 1.

If tn = 20, d = -1 and n = 16, then the first term is?

Options:

A. 36

B.-35

C. 35

D. 5

#### Solution:

tn = a + (n – 1)d

20 = a + (16 – 1) * (-1)

20 = a + (15) * (-1)

20 = a – 15

-a = – 15 – 20

– a = – 35

a = 35

### Question 2.

What is the position of 62 in the following series 2, 5, 8, ….?

Options:

A. 21

B. 22

C. 23

D. 20

#### Solution:

In the given series,

a = 2

d = 3

tn = 62

tn = a + (n – 1)d

62 = 2 + (n -1) * 3

62 = 2 + 3n – 3

62 – 2 + 3 = 3n

63 = 3n

n = 21

### Question 3.

Find the 11th term of the series 1, 3.5, 6, 8.5, ……

Options:

A. 25

B. 26

C. 23.5

D. 24.5

#### Solution:

In the given series a = 1, d = 2.5

n =11

tn = a + (n – 1)d

tn = 1 + (11 – 1) * 2.5

tn = 1 + 10 * 2.5

tn = 1+ 25

tn = 26

## Type 2: Number of terms in the series: n=  [(l-a)/d]+1

### Question 1.

Find the number of terms in the series 2, 4, 6, . . .98

Options:

A. 50

B. 49

C.51

D. 48

#### Solution

We know that,

n= [(l-a)/d]+ 1

a (first term) = 2
l (last term) = 98
d (common difference) = 2

n= (98-2)/2 + 1

n = (96/2) + 1

n = 48 + 1

n = 49

### Question 2.

Find the last number of the series where, a = 3, d = 5, and number of terms are 45

Options:

A. 225

B.223

C.230

D. 222

#### Solution

According to the question,

a = 3, d = 5, n = 45

n= [(l-a)/d]  + 1
45 =  [(l-3)/5]+ 1
45-1 = (l-3)/5

44*5 = l – 3

220 = l-3

l=223

Correct option: B

### Question 3.

Find the total odd numbers  in the series 3….. 99.

Options:

A. 49

B.51

C. 50

#### Solution:

The common difference is 2 because it is an odd number series.

We know that,

n= (l-a)/d + 1

a = 3
l = 99
d (common difference) = 2

n= (99-3)/2  + 1

n = 96/2 + 1

n = 48 + 1

n = 49

## Type 3: Sum of first ‘n’ terms of the series: Sn = n/2*[2a + (n − 1) d] or n/2*(a+l)

### Question 1.

Sanjana has one paisa on 1st of January with her. She got 2 paisa on the 2nd January and 3 paisa on 3rd of January. In similar fashion, what will be her saving after 365 days?

Options:

A. 66700 paisa

B. 66795 paisa

C. 65795 paisa

D. 66790 paisa

#### Solution:

According to the question, the series formed is 1,2,3,4,5,……

Therefore, a = 1 and d = 1

Sn = n/2*[2a + (n − 1) d]
n = 365

S365 =  365/2*[2 * 1 + (365 − 1) * 1]

S365 = 66795 paisa

### Question 2.

Find the sum of the series 62 + 60 + 58 …….. + 30

Options:

A. 728

B.780

C.719

D. 782

#### Solution:

In the given series a = 62, d = -2, tn = 30

tn = a + (n – 1)d

30= 62 + (n – 1) * -2

30 = 62 – 2n + 2

2n = 62-30+2

34 = 2n

n = 17

Sn = n/2(a+l)

S17 = 17/2 (62 + 30)

S17 =  17/2(92)

S17 = 782

### Question 3.

If the sum of 5 consecutive numbers is 100 then find out the first number.

Options:

A. 20

B. 25

C. 18

D. 16

#### Solution:

It is clear from the question that d = 1 because there are 5 consecutive numbers

n = 5

Sn= 100

Let the first number ‘a’ be x

Sn = n/2[2a + (n − 1) d]

100 = 5/2 [2a + (5 − 1) * 1]

100 =  5/2[2a + (4) * 1]

100*2/5 =  [2a + 4]

40 = 2a + 4

a = 18

So, the first number is 18 and the complete series is 18, 19, 20, 21, and 22.

## Type 4: Arithmetic mean of the series: b =  (a + c)/2

### Question 1.

Find the arithmetic mean of first five prime numbers.

Options:

A. 4.5

B. 6.5

C. 5.6

D. 6.4

#### Solution:

We know that, arithemetic mean of a,b,c is

b =  (a + c)/2

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

Therefore, there arithmetic mean (AM) =(2+3+5+7+11)/5  = 5.6

### Question 2.

Find the arithmetic mean of first 11 whole numbers.

Options:

A. 4

B. 6

C.5

D.6

#### Solution:

We know that,

b =  1/2(a + c)

Therefore, arithmetic mean (AM) = 0+1+2+3+4+5+6+7+8+9+10)/11= 5

### Question 3.

Find the arithmetic mean of series 2, 4, 6 , 8….100

Options:

A. 51

B.50

C.25

D.49

#### Solution:

In the given series, a = 2, d = 2, l = 100

To find the total number of term we can use the formula

n= [(l-a)/d]  + 1

n =  [(100-2)/2]+ 1

n = 50

Now, to find the sum of the series we can use the formula Sn = [2a + (n − 1) d] OR   (a+l)

Sn = n/2 (a+l)

Sn =  n/2(a+l)

S50 =  50/2(2+l00)

S50 = 25 (102)

S50 = 2550

Now, AM = 2550/50 = 51