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Special Series Questions

Learn about special series questions, their formula, and practice with related maths and statistics questions in this study material.

Arithmetic progression refers to the sequence of numbers in mathematics where we see the difference between two consecutive terms, as well as the difference being completely constant. And these differences observed are known as common differences. Also, this property is known as an arithmetic sequence. The first term that appears in this sequence is known as the starting term of this sequence, which is described as “a” in the arithmetic sequence. Similarly, the difference between two terms found in this sequence is known as the parallelogram, which we refer to as “d”. Here the nth term of an arithmetic progression is shown here, which are as follows:

an = a + (n-1) d

Better Example to Understand for arithmetic Progression-

 

Example

Common Difference

d

a

10,20,30,40….

If the value of d is greater than zero, then it will show positive nature.

10

10

50,40,30,20…

If the value of d is less than zero, then it will show negative nature.

-10

50

2,2,2,2..

If the value of d is equal to zero, then it will show zero.

0

2

 A geometric progression refers to a sequence of numbers where each subsequent term of the sequence by multiplying or dividing the preceding number by a fixed number represents its subsequent number or the previous number, which is known as a geometric sequence. The first term of the sequence is known as a, which is called the starting term. Also, the ratio of the next term of the sequence of the term is known as the common ratio, which is represented as ‘r’. Thus look at the nth term of the geometric progression.

If the series…a, ar, ar2, ar3, …, arn

 

an = a. rn-1

Solved examples of special series

1. What is the tenth term in the 2 + 4 + 6 + 8 +… series?

Solution:

a = 2 and d = 2 in the given sequence

As a result, the values will be entered into the formula for the nth term of the series.

an = a + (n – 1) d

Here n = 10 as we have to find the 10th term.

a10 = 2 + (10 – 1) 2

= 2 + 18

= 20

So the 10th term of the series will be 20

2. What is the sequence’s common ratio?

3, 6, 12, 24, …

Solution:

Given

a1 = 3

a2 = 6

a3 =12

a4 = 24

r= a2a1 = 63

 =a3a2=126 

As a result, the common ratio is 2.

3. Find the sum of the given arithmetic progression

2 + 5 + 8 + 11 + 14

Solution:

Given

a = 2 

d = 3

n = 5 

Now we will put the values in the formula

an = a + (n – 1) d

The total of the series would be if we knew the last term ‘l’ of the series instead of the common difference ‘d’.

Sn = n2 ( a + l )

Consider the following equation: 2 + 5 + 8 + 11 + 14

We’ll now calculate the sum of the series with a=2 and l=14.

Sn = n2 ( a + l )

= 52 ( 2 + 14 )

= 40

4. Five integers between 3 and 21 must be inserted in such a way that the resulting sequence is an AP

Solution:

Let A1, A2, A3, A4, and A5 be the five numbers between 3 and 21 that have the following properties:

3, A1, A2, A3, A4, A5, 21

The sequence is in AP 

Here, a = 3, b = 21, n = 7.

Therefore,

21 = 3 + (7 –1) d

18 = 6d

d = 3

Thus, A 1 = a + d = 3 + 3 = 6;

A2 = a + 2d = 3 + 2 × 3 = 9;

A3 = a + 3d = 3 + 3 × 3 = 12;

A4 = a + 4d = 3 + 4 × 3 = 15;

A5 = a + 5d = 3 + 5 × 3 = 18;

As a result, the five digits from 3 to 21 are 6, 9, 12, 15, and 18.

5. What is the 6th term in the 2 + 4 + 8 + 16 +… series?

In the given series, a = 2 and r = 2 are the correct answers.

As a result, the values will be entered into the formula for the nth term of the series.

an = a.rn-1

We use n = 6 since we need to find the sixth term.

a6 = 2.26-1

= 2 .32

= 64

As a result, the series’ sixth term will be 32.

6. What does the Geometric Mean of 1, 3, 9, 27, and 81 look like?

Solution:

Given

a1 = 1

a2 = 3

a3 =9

a4 = 27

a5 = 81

We’ll start by multiplying the integers

1 × 3 × 9 × 27 × 81 = 59049

Take the 5th root (that is, the nth root, when n = 5) after that.

5√59049 = 9

Geometric Mean = 5√ (1 × 3 × 9 × 27 × 81) = 9

7. If Radha saves some money in her piggy bank every month, how much money will she have in her piggy bank after 12 months if the money is in the following order: 100, 150, 200, 250,….

Solution: The following is the sequence:

100, 150, 200, 250, …

a = 100 

d = 50 

n = 12 people 

We’ll now enter the values into the formula.

Sn = n2 2a + (n- 1 ) d 

Sn = 122 2. 100 + (12 – 1 ) 50

Sn = 6 200 + (11 ) 50

Sn = 4,500

So she has Rs. 4,500 in her bank account after a year.

Conclusion

Arithmetic progression is the main part of a series. It refers to a difference between terms as a constant value in consecutive terms. It is mainly used for making patterns. We can see its use in our everyday life. At the same time, it helps in making tasks like determining the number of audience members and crowd easier. Apart from this, it helps in calculating the estimated income in any big company.

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