# How To Solve Arithmetic Progression Questions Quickly ## How To Solve Arithmetic Progression Questions Quickly

Arithmetic progression is a sequence of terms such that the difference between the consecutive terms is a constant value . Or we can say that a succession of terms or numbers formed and arranged in a definite order according to a definite rule is called a Arithmetic Progression.

### 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:    We know that,  tn = a + (n – 1)d

20 = a + (16 – 1) * (-1)

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

20 = a – 15

-a = – 15 – 20

– a = – 35

a = 35

Correct option: C

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

Correct option: A

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

Correct option: B

### Type 2: Number of terms in the series: $n = \left [ \frac{l – a}{d} \right ] + 1$

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, $n = \left [ \frac{l – a}{d} \right ] + 1$

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

Correct option: B

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

$n = \left [ \frac{l – a}{d} \right ] + 1$

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

D. 60

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

We know that,

$n = \left [ \frac{l – a}{d} \right ] + 1$

a = 3

l = 99

d (common difference) = 2

n= (99-3)/2  + 1

n = 96/2 + 1

n = 48 + 1

n = 49

Correct option: A

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

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

$S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]$

n = 365

S365 =  365/2*[2 * 1 + (365 − 1) * 1]

S365 = 66795 paisa

Correct option: C

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

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

S17 = 17/2 (62 + 30)

S17 =  17/2(92)

S17 = 782

Correct option: D

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

$S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]$

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.

Correct option: C

### Type 4: Arithmetic mean of the series: $b = \frac{(a + c)}{2}$

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

Correct option: C

Question 2. Find the arithmetic mean of first 11 whole numbers.

Options:

A. 4

B. 6

C.5

D.6

Solution:    We know that,

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

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

Correct option: C

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

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

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   n/2 (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

Correct option: A