Sequence Questions

Sequence questions are given here to help the students to get a thorough understanding of the sequences in Maths. All these questions contain detailed solutions and formulas required to reach the final answer. In this article, you will get sequence questions based on the latest NCERT curriculum since it is one of the important topics in Class 11 Maths.

What is a Sequence in Maths?

In mathematics, a sequence is a list of numbers in a specific order, and the numbers in such a list are called the terms of the sequence. A sequence is either finite or infinite, depending upon the number of terms in a sequence. Thus, a sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence, i.e. the number of terms is not finite.

Also, check: Sequences.

Sequence Questions and Answers

1. Consider a sequence: 1, 10, 9, x, 25, 26, 49. Find x.

Solution:

The given sequence is a combination of two sequences:

Sequence – I: 1, 9, 25, 49 and

Sequence – II.: 10, x, 26

Here,

The pattern in I is +8, +16, +24.

The pattern in II is 3² + 1, 4² + 1, 5² + 1.

i.e. 10, 17, 26

So, 17 must be the missing number.

Therefore, the value of x is 17.

2. What is the 20th term of the sequence defined by a_n = (n – 1) (2 – n) (3 + n)?

Solution:

Given,

a_n = (n – 1)(2 – n)(3 + n)

Substituting n = 20, we get;

a₂₀ = (20 – 1)(2 – 20)(3 + 20)

= 19 × (-18) × 23

= -7866

Therefore, the 20th term of the given sequence is -7866.

3. Write the first five terms of the sequence for which a₁ = 3, a_n = 3a_{n – 1} + 2 for all n > 1.

Solution:

Given,

a₁ = 3

a_n = 3 a_n-1 + 2; n > 1

When n = 2,

a₂ = 3a₁ + 2 = 3(3) + 2 = 11

When n = 3,

a₃ = 3a₂ + 3 = 3(11) + 2 = 35

When n = 4,

a₄ = 3a₃ + 2 = 3(35) + 2 = 107

When n = 5,

a₅ = 3a₄ + 2 = 3(107) + 2 = 323

Therefore, the first five terms of the given sequence are 3, 11, 35, 107, 323.

4. What is the sum of odd integers from 1 to 1001?

Solution:

Odd integers from 1 to 1001 are: 1, 3, 5, …, 999, 1001

This is an arithmetic sequence with the first term, a = 1 and the common difference, d = 2, last term, l = 1001.

Let n be the number of terms in the sequence.

So, a_n = l

a + (n – 1)d = 1001

1 + (n – 1)2 = 1001

(n – 1)2 = 1000

(n – 1) = 500

n = 501

Sum of the sequence = (n/2)(a + l)

= (501/2) (1 + 1001)

= (501 × 1002)/2

= 501 × 501

= 251001

Hence, the sum of the odd integers from 1 to 1001 is 251001.

5. What is the 80th term of the sequence 2, 3, 5, 8, 12, 17, 23,…?

Solution:

Given sequence:

2, 3, 5, 8, 12, 17, 23,…

The differences between two consecutive terms of this sequence are 1, 2, 3, 4, 5, 6,…

Thus, it is neither arithmetic nor geometric sequence.

So, let’s find the general term for this sequence.

The second differences between the terms, i.e. for 2, 3, 4,… are:

1, 1, 1, 1, 1,…

A constant sequence is obtained after taking differences twice, so we can match the nth term of the sequence with a quadratic formula. Thus, based on the initial term of each of the sequences we can write it as:

a_n = (2/0!) + (1/1!) (n – 1) + (1/2!)(n – 1)(n – 2)

= 2 + (n – 1) + (1/2)(n – 1)(n – 2)

= 2 + n – 1 + (1/2) (n² – 3n + 2)

= n + 1 + (1/2) (n² – 3n + 2)

= (2n + 2 + n² – 3n + 2)/2

= (n² – n + 4)/2

Thus, the nth term of the given sequence is a_n = (n² – n + 4)/2

Substituting n =80, we get;

a₈₀ = [(80)² – 80 + 4]/2

= (6400 – 76)/2

= 6324/2

= 3162

Therefore, the 80th term of the given sequence is 3162.

6. Find the sum of all numbers between 200 and 400, which are divisible by 7.

Solution:

The numbers lying between 200 and 400, which are divisible by 7, are:

203, 210, 217, ­­­­­­­­… 399

This sequence is an AP with the first term, a = 203, last term, l = 399 and the common difference, d = 7.

Let n be the number of terms of the sequence.

∴ a_n = l

a + (n –1) d = 399

203 + (n – 1) 7 = 399

(n – 1)7 = 399 – 203

⇒ 7(n – 1) = 196

⇒ n – 1 = 196/7 = 28

⇒ n = 29

Sum of the sequence = (n/2)(a + l)

= (29/2) (203 + 399)

= (29/2) (602)

= 29 × 301

= 8729

Thus, the required sum is 8729.

7. The sum of some terms of a geometric sequence is 315, whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Solution:

Let 315 be the sum of n terms of a geometric sequence.

Given,

First term = a = 5

Common ratio = r = 2

We know that,

Sn = a(rⁿ – 1)/(r – 1); r > 1

315 = [5(2ⁿ – 1)]/ (2 – 1) {from the given Sn = 315}

315 = 5(2ⁿ – 1)

⇒ 2ⁿ – 1 = 315/5

⇒ 2ⁿ – 1 = 63

⇒ 2ⁿ = 64

⇒ 2ⁿ = 2⁶

⇒ n = 6

Thus, the sum of 6 terms is 315.

We know that the nth term of the geometric sequence = ar^n-1

6th term of the given sequence = (5)×(2)^6-1

= 5 × 2⁵

= 5 × 32

= 160

Therefore, the number of terms of the sequence is 6, and the last term, i.e. the 6th term of the sequence, is 160.

8. Find the sum of the first five terms of the sequence given by the recurrence relation a_n+2 = (a_n – a_n+1)² – 10 and a₁ = 5, a₂ = 1.

Solution:

Given,

a₁ = 5

a₂ = 1

And

a_n+2 = (a_n – a_n+1)² – 10

Let us find the next three terms of the sequence.

a₃ = a_{(1 + 2)}

= (a₁ – a₂)² – 10

= (5 – 1)² – 10

= 16 – 10

= 6

a₄ = (a₂ – a₃)² – 10

= (1 – 6)² – 10

= 25 – 10

= 15

a₅ = (a₃ – a₄)² – 10

= (6 – 15)² – 10

= 81 – 10

= 71

Thus, the first 5 terms of the sequence are 5, 1, 6, 15, 71.

Sum of the sequence = 5 + 1 + 6 + 15 + 71 = 98

Therefore, the sum of the first five terms of the sequence is 98.

Practice Questions on Sequences

1. If the nth term of the sequence is 4n – 3, find the 17th and 24th terms of the sequence.
2. Write the first four terms in each of the following sequences defined by a_n = 2n + 5.
3. The product of three numbers of an arithmetic sequence is 224, and the largest number is 7 times the smallest. Find the numbers.
4. Find the sum of the sequence 7, 77, 777, 7777, … to n terms.
5. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
6. Find the sum to n terms of the sequence whose nth term is n(n + 3).