Practice Questions on Arithmetic Series and Sequence

Last Updated : 23 Sep, 2025

A sequence of numbers is called an arithmetic progression or sequence if the difference between any two consecutive terms is always the same.

For example, 2, 4, 6, 8, 10 is an AP because the difference between any two consecutive terms in the series (common difference) is the same:

4 – 2 = 6 – 4 = 8 – 6 = 10 – 8 = 2. 

The first term of AP is denoted as "a". In a series, it can be described as follows:

a, a + d, a + 2d, a + 3d, a + 4d, . . . , a + (n – 1) d

The sum of the terms of an arithmetic sequence is called an arithmetic series.

Solved Practice Problems on Arithmetic Series and Sequence

Question 1: Find the common difference of the sequence 12, 27, 42, 57, 72, . . .

Solution:

Given sequence is 12, 27, 42, 57, 72, . . .

Formula Used: Common difference (d) = Difference between any two consecutive terms

So, d = 27 - 12 = 42 - 27 = 57 - 42 = 72 - 57 = 15

Required common difference is 15.

Question 2: Find the 12th term of the sequence 8, 12, 16, 20, . . .

Solution:

Given series is 8, 12, 16, 20, . . .

Formula used: nth term (Tn) = a + (n − 1) × d

Here, a = 8

d = 12 - 8 = 4

And, n = 12

We know that, nth term (Tn) = a + (n − 1) × d

So, T12 = 8 + (12 - 1)4 = 8 + 44 = 52

∴ Required value is 52

Question 3: Find the number of terms in the progression 15, 30, 45, 60, . . . , 480.

Solution:

We have the series 15, 30, 45, 60, . . . , 480.

Formula Used: n = {(last term - first term)/d} + 1, where d = Difference between any two consecutive terms of Series

d = 30 - 15 = 15

According to the formula, we have

n = {(480 - 15)/15} + 1 = 31 + 1 = 32

∴ Required number of terms is 32.

Question 4: Find the 15th terms of the sequence 122, 133, 144, . . .

Solution:

Given series is 122, 133, 144, . . .

Formula used: Tn = a + (n - 1)d

Calculation:

In the given series, d = 133 - 122 = 11

a = 122

According to the formula, we have

T15 = 122 + (15 - 1) × 11 = 122 + 14 × 11 = 122 + 154 = 276

∴ Required number of term is 276.

Question 5: Find the sum of the progression 14, 50, 86, 122, 158, 194, 230, 266, 302.

Solution:

Given series is 14, 50, 86, 122, 18, 194, 230, 266, 302.

Formula used: Sn = n/2(first term + last term)

We have, n = 9

First term = 14, last term = 302

According to the formula, we have

Sum of the series = (9/2)(14 + 302) = (9/2) × 316 = 1422

∴ Sum of the series is 1422.

Question 6: If the sum of n terms of an A.P. is 12000 and the first and last terms are 75 and 125 then find the value of n.

Solution:

Given: Sum of the A. P. is 12000

  • First-term = 75
  • Last term = 125

Formula used: Sum of n terms = (n/2)(first term + last term)

According to the formula, we have

12000 = (n/2)(75 + 125)

So, 100n = 12000

Hence, n = 120

∴ Required value is 120

Question 7: Find the sum of the series using 20 terms from sequence 20, 38, 56, 74, . . .

Given:

Given series is 20, 38, 56, 74, . . .till 20 terms.

Formula used: Sn = (n/2){2a + (n - 1)d}

In the given series, a = 20

d = 38 - 20 = 18

n = 20

Then according to the formula, we have

Sum of the series = (20/2){(2 × 20 + (20 - 1) × 18} = 10(40 + 342) = 3820

∴ Sum of the series is 3820

Question 8: Find the sum of the series -5, -10, -15, -20, . . . , -855

Solution:

Given: series is -5, -10, -15, -20, . . . , -855

Here a = -5

Last term = -855

Here d = -10 + 5 = -5

Let the number of terms in the series be n

Then, last term = -5 + (n - 1)(-5)

So, -855 = -5 - (n - 1)5

Then, n = 171

We know that, sum of the series = (n/2)(first term + last term) = (171/2)(-5 - 855) = 171 × -430 = -73530

∴ Required value is -73530

Question 9: Find the sum of the first 1000 odd numbers.

Solution:

Required series is 1, 3, 5, 7, . . .

Formula used: Sn = (n/2){2a + (n - 1)d}

Here, a = 1

d = 3 - 1 = 2

And, n = 1000

So, sum of the series = (1000/2){2 + (1000 - 1)2} = 500 × 2000 = 10, 00, 000.

∴ Required value is 1000000.

Question 10: Find the sum of the first 30 multiples of 5.

Solution:

Formula used: Sn = (n/2)[2a + (n − 1) × d] 

Calculation:

Required series is 5, 10, 15, 20, . . .

Here, a = 5

d = 5

So, Sn = (n/2)[2a + (n − 1) × d] 

= (30/2) (10 + 29 × 5) = 15 × 155 = 2325

∴ Required value is 2325

Unsolved Practice Problems on Arithmetic Series and Sequence

Problem 1: Find the sum of the first 20 terms of the arithmetic series 5, 8, 11, 14, . . ..

Problem 2: The 7th term of an arithmetic series is 15 and 12th term is 30. Find the first term and common difference.

Problem 3: If the sum of the first 10 terms of an arithmetic series is 155 and first term is 5 find the common difference.

Problem 4: Determine the number of terms in the arithmetic series 3, 7, 11, . . ., 103.

Problem 5: The sum of the first n terms of an arithmetic series is given by Sn = 3n2+5n. Find the first term and the common difference.

Problem 6: Find the 15th term of the arithmetic series where the 5th term is 7 and 10th term is 17.

Problem 7: The sum of the first 15 terms of an arithmetic series is 300 and sum of the first 10 terms is 150. Find the first term and the common difference.

Problem 8: An arithmetic series has a common difference of 4. If the 10th term is 35 find the first term and sum of the first 12 terms.

Problem 9: If the sum of the first 20 terms of an arithmetic series is 400 and common difference is 2 find the first term.

Problem 10: The 4th term of an arithmetic series is twice the first term and 6th term is 18. Find the first term, the common difference and sum of the first 10 terms.

Answer Key

  1. 670
  2. First term: -3 , Common difference: 3
  3. Common difference: 7/3
  4. 26
  5. First term: 8, Common difference: 14.
  6. 27
  7. First term: 6, Common difference: 2
  8. First term: -1, Sum of first 12 terms: 252
  9. First term: 1
  10. First term: 6, Common difference: 2, Sum of first 10 terms: 150
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