Mathematics Part I Solutions Solutions for Class 10 Math Chapter 1 Arithemetic Sequences are provided here with simple step-by-step explanations. These solutions for Arithemetic Sequences are extremely popular among Class 10 students for Math Arithemetic Sequences Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 10:

Question 1:

The perimeters of squares with each side of length 1 centimetre, 2 centimetres, 3 centimetres and so on. Also, the areas of these squares.

Answer:

Perimeter of the square = 4 × side 

Perimeter of the square with side 1 cm = 4 × 1 cm = 4 cm

Perimeter of the square with side 2 cm = 4 × 2 cm = 8 cm

Perimeter of the square with side 3 cm = 4 × 3 cm = 12 cm

Perimeter of the square with side 4 cm = 4 × 4 cm = 16 cm

...

So, we get the number sequence as:

4, 8, 12, 16, …

Area of the square = (side)2 

Area of the square with side 1 cm = (1 cm)2 = 1 cm2

Area of the square with side 2 cm = (2 cm)2 = 4 cm2

Area of the square with side 3 cm = (3 cm)2 = 9 cm2

Area of the square with side 4 cm = (4 cm)2 = 16 cm2

.

So, we get the number sequence as:

1, 4, 9, 16, …

Page No 10:

Question 2:

The sum of the interior angles of polygons of sides 3, 4, 5 and so on. And the sum of their exterior angles.

Answer:

Sum of the interior angles of a polygon with n sides = (n − 2) × 180°

Sum of the interior angles of a polygon with 3 sides = (3 − 2) × 180° = 1 × 180° = 180°

Sum of the interior angles of a polygon with 4 sides = (4 − 2) × 180° = 2 × 180° = 360°

Sum of the interior angles of a polygon with 5 sides = (5 − 2) × 180° = 3 × 180° = 540°

Sum of the interior angles of a polygon with 6 sides = (6 − 2) × 180° = 4 × 180° = 720°

So, we get the number sequence as:

180, 360, 540, 720, …

The sum of the exterior angles of a polygon with any number of sides is 360°

So, we get the number sequence as:

360, 360, 360, 360, …

Page No 10:

Question 3:

Multiples of 3

Numbers which leave a remainder 1 on dividing by 3

Number which leave a remainder 2 on dividing by 3

Answer:

Multiples of 3 are 3, 6, 9, 12, …

So, we get the number sequence as:

3, 6, 9, 12, …

If we add 1 to the multiples of 3, we get numbers which leave their remainder as 1 on being divided by 3. These numbers are (3 + 1), (6 + 1), (9 + 1), (12 + 1), …

So, we get the number sequence as:

4, 7, 10, 13, …

If we add 2 to the multiples of 3, we get numbers which leave their remainder as 2 on being divided by 3.These numbers are (3 + 2), (6 + 2), (9 + 2), (12 + 2), …

So, we get the number sequence as:

5, 8, 11, 14, …

Page No 10:

Question 4:

Natural numbers which end in 1 or 6, written in order. Can you describe this sequence in any other way?

Answer:

The sequence of natural numbers which end in 1 or 6 is 1, 6, 11, 16, 21, 26, …

This sequence can also be described as:

First number in the given sequence is 1 and the succeeding numbers are obtained by adding 5. 



Page No 12:

Question 1:

The multiples of 2, written in order given 2, 4, 6, 8, …… Is this an arithmetic sequence? What about the powers 2, 4, 8, 16, …… of 2?

Answer:

An arithmetic sequence is a sequence which starts with any number and proceeds by addition of a number again and again.

The given sequence is 2, 4, 6, 8, …

It starts with 2 and proceeds by addition of 2.

So, this is an arithmetic sequence.

The powers 2, 4, 8, 16, … of 2 start with 2.

However, the sequence proceeds by addition of 2, 4, 8, …

So, this is not an arithmetic sequence.

Page No 12:

Question 2:

Dividing the natural numbers by 2, we get Is this an arithmetic sequence?

Answer:

An arithmetic sequence is a sequence which starts with any number and proceeds by addition of a number again and again.

The given sequence is

It starts with and proceeds by addition of .

So, this is an arithmetic sequence.

Page No 12:

Question 3:

In the arithmetic sequence got by adding to itself repeatedly, does the number 10 occur anywhere? What about 11?

Answer:

First number =

Second number =

Third number =

Fourth number =

Fifth number =

Sixth number =

Seventh number =

Eighth number =

Since the consecutive natural numbers, 1 and 2, occur in the given sequence, all the natural numbers will occur in the given sequence.

10 and 11 will also occur in the given sequence.

Page No 12:

Question 4:

The reciprocals of the natural in order give Is it an arithmetic sequence?

Answer:

An arithmetic sequence is a sequence which starts with any number and proceeds by addition of a number again and again.

The given sequencestarts with 1.

However, the sequence proceeds by addition of

So, this is not an arithmetic sequence.

Page No 12:

Question 5:

Write down the sequence of differences (subtracting the larger form the smaller) of consecutive perfect squares. Is it and arithmetic sequence?

Answer:

Consecutive perfect squares are 1, 4, 9, 16, 25, 36, …

The sequence formed by the differences of consecutive perfect square is (4 1), (9 4), (16 9), (25 16), …

Or 3, 5, 7, 9, …

The sequence 3, 5, 7, 9, … starts with 3 and proceeds by addition of 2.

We know that an arithmetic sequence is a sequence which starts with any number and proceeds by addition of a number again and again.

So, this is an arithmetic sequence.



Page No 14:

Question 1:

In each of the arithmetic sequences below, some numbers are missing. Their positions are marked with a. Find these numbers.

Answer:

(1)

First number = 24

Second number = 42

Let the third number be x and the fourth number be y

Thus, the arithmetic sequence is 24, 42, x, y, …

42 − 24 = x − 42 

x = 42 − 24 + 42

x = 60

Also, x − 42 = yx 

60 − 42 = y − 60 

y = 60 + 60 − 42 

y = 78

Therefore, the third and the fourth numbers are 60 and 78 respectively.

Thus, the arithmetic sequence is:


 

(2)

Second number = 24

Third number = 42

Let the first number be x and the fourth number be y

Thus, the arithmetic sequence is x, 24, 42, y, …

24 − x = 42 − 24 

x = 24 − 42 + 24 

x = 6 

Also, 42 − 24 = y − 42 

y = 42 − 24 + 42 

y = 60

Therefore, the first and the fourth numbers are 6 and 60 respectively.

Thus, the arithmetic sequence is:


 

(3)

Third number = 24

Fourth number = 42

Let the first number be x and the second number be y.

Thus, the arithmetic sequence is x, y, 24, 42, …

24 − y = 42 − 24

y = 24 − 42 + 24

y = 6

Also, yx = 24 − y 

6 − x = 24 − 6

x = 6 − 24 + 6

x = 12

Therefore, the first and the second numbers are 12 and 6 respectively.

Thus, the arithmetic sequence is:


 

(4)

First number = 24

Third number = 42

Let the second number be x and the fourth number be y.

Thus, the arithmetic sequence is 24, x, 42, y, …

x − 24 = 42 − x 

x + x = 42 + 24

2x = 66

x = 33

Also, 42 − x = y − 42 

42 − 33 = y − 42

y = 42 − 33 + 42

y = 51 

Therefore, the second and the fourth numbers are 33 and 51 respectively.

Thus, the arithmetic sequence is:


 

(5)

Second number = 24

Fourth number = 42

Let the first number be x and the third number be y.

Thus, the arithmetic sequence is x, 24, y, 42, …

y − 24 = 42 − y

y + y = 42 + 24

2y = 66

y = 33

Also, 24 − x = y − 24 

24 − x = 33 − 24

x = 24 − 33 + 24 

x = 15 

Therefore, the first and the third numbers are 15 and 33 respectively.

Thus, the arithmetic sequence is:


 

(6)

First number = 24

Fourth number = 42

Let the second number be x and the third number be y.

Thus, the arithmetic sequence is 24, x, y, 42, …

x − 24 = yx 

2xy − 24 = 0 … (1)

Also, yx = 42 − y

2yx − 42 = 0 … (2)

Multiplying equation (2) by 2:

4y − 2x − 84 = 0 … (3)

On solving equations (1) and (3):

3y − 108 = 0

y = 36

From equation (1), x = 30 

Therefore, the second and the third numbers are 30 and 36 respectively.

Thus, the arithmetic sequence is:

Page No 14:

Question 2:

The second and fourth numbers in an arithmetic sequences are 8 and 2. Find the first and the third numbers.

Answer:

Second number in an arithmetic sequence = 8

Fourth number in an arithmetic sequence = 2

Let the first and the third number in the arithmetic sequence be x and y respectively.

Thus, the arithmetic sequence is x, 8, y, 2, …

y − 8 = 2 − y 

y + y = 2 + 8

2y = 10

y = 5

Also, 8 − x = y − 8

8 − x = 5 − 8

x = 8 − 5 + 8

x = 11

Therefore, the first and the third number in the arithmetic sequence is 11 and 5 respectively.



Page No 15:

Question 1:

The second number of and arithmetic sequence is 5. Find the sum of its first and third numbers.

Answer:

Second number in the arithmetic sequence = 5

Let the first and the third number in the arithmetic sequence be x and y respectively.

Thus, the arithmetic sequence is x, 5, y, …

5x = y − 5 

x + y = 5 + 5

x + y = 10

Therefore, the sum of the first and the third numbers of the arithmetic sequence is 10.

Page No 15:

Question 2:

Find the algebraic expression to compute the third number of an arithmetic sequence, using the first and the second numbers.

Answer:

Let the first, second and third number in the arithmetic sequence be x, y and z respectively.

Thus, the arithmetic sequence is x, y, z, …

yx = zy 

z = yx + y

z = 2yx 

Therefore, the algebraic expression to compute the third number of the arithmetic sequence using the first and the second numbers is z = 2yx, where x, y and z are respectively the first, second and third numbers of the arithmetic sequence.



Page No 17:

Question 1:

The first term of an arithmetic sequence is 7 and the common difference is −2. What is its 12th term?

Answer:

1st term of the arithmetic sequence = 7

Common difference = 2

To get the 12th term from the 1st term, we must add the common difference 11 times to the 1st term.

12th term = 7 + 11(2) = 7 − 22 = 15

Page No 17:

Question 2:

The 3rd term of an arithmetic sequence is 10 and its 8th term is 25. What is its 4th term? And the 13th term?

Answer:

3rd term of the arithmetic sequence = 10

8th term of the arithmetic sequence = 25

To get the 8th term from the 3rd term, we must add the common difference 5 times to the 3rd term.

The number to be added is 25 − 10 = 15

Common difference = = 3

To get the 4th term from the 3rd term, we must add the common difference once to the 3rd term.

4th term = 10 + 1 × 3 = 10 + 3 = 13

To get the 13th term from the 3rd term, we must add the common difference ten times to the 3rd term.

13th term = 10 + 10(3) = 10 + 30 = 40

Page No 17:

Question 3:

The 5th term of an arithmetic sequence is 11 and its 12th term is 32. What are its first three terms?

Answer:

5th term of the arithmetic sequence = 11

12th term of the arithmetic sequence = 32

To get the 12th term from the 5th term, we must add the common difference seven times to the 5th term.

The number to be added is 32 − 11 = 21

Common difference = = 3

We can get the 5th term from the 1st term by adding the common difference four times to the 1st term.

So, in this sequence, the 5th term, i.e., 11 is received by adding 4 × 3 = 12 to the 1st term.

1st term = 11 − 12 = 1

Now, the 1st term is 1 and the common difference is 3.

2nd term = 1 + 3 = 2

3rd term = 2 + 3 = 5

So, the first three terms of the given arithmetic sequence are 1, 2 and 5.

Page No 17:

Question 4:

The 5th term of an arithmetic sequence is 9 and its 9th term is 5. What is its common difference? What is the 14th term?

Answer:

5th term of the arithmetic sequence = 9

9th term of the arithmetic sequence = 5

To get the 9th term from the 5th term, we must add the common difference four times to the 5th term.

The number to be added is 5 − 9 = 4

Common difference = = 1

To get the 14th term from the 5th term, we must add the common difference nine times to the 5th term.

14th term = 9 + 9(1) = 9 − 9 = 0

Page No 17:

Question 5:

The common difference of an arithmetic sequence is −1 and its 4th term is 7. What is its 7th term? And the 11th term?

Answer:

4th term of the arithmetic sequence = 7

Common difference = 1

To get the 7th term from the 4th term, we must add the common difference three times to the 4th term.

7th term = 7 + 3(1) = 7 − 3 = 4

To get the 11th term from the 4th term, we must add the common difference seven times to the 4th term.

11th term = 7 + 7(1) = 7 − 7 = 0

Page No 17:

Question 6:

How many three digit numbers leave a remainder 3 on division by 4?

Answer:

1st three-digit number which leaves the remainder 3 when divided by 4 = 103

2nd three-digit number which leaves the remainder 3 when divided by 4 = 107

3rd three-digit number which leaves the remainder 3 when divided by 4 = 111

nth three-digit number which leaves the remainder 3 when divided by 4 = 999

So, the arithmetic sequence is 103, 107, 111, … , 999

Common difference = 107 − 103 = 4

To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n 1) times to the 1st three-digit number.

999 = 103 + (n − 1) × 4

999 − 103 = 4n − 4 

4n − 4 = 896 

4n = 900 

n = 225

Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.

Page No 17:

Question 7:

Write down the sequence of natural numbers which leave a remainder 3 on division by 6. What is the 10th term of this sequence? How many terms of this sequence are between 100 and 400?

Answer:

(1) 

1st natural number which leaves the remainder 3 when divided by 6 = 9

2nd natural number which leaves the remainder 3 when division by 6 = 15

3rd natural number which leaves the remainder 3 when divided by 6 = = 21

So, the arithmetic sequence is 9, 15, 21, …


 

(2)

Common difference = 15 − 9 = 6

To get the 10th natural number from the 1st natural number, we must add the common difference nine times to the 1st natural number.

10th term = 9 + 9 × 6 = 9 + 54 = 63


 

(3)

1st natural number between 100 and 400 which leaves the remainder 3 when divided by 6 = 105

2nd natural number between 100 and 400 which leaves the remainder 3 when divided by 6 = 111

3rd natural number between 100 and 400 which leaves the remainder 3 when divided by 6 = 117

nth natural number between 100 and 400 which leaves a remainder 3 on division by 6 = 399

So, the arithmetic sequence is 105, 111, 117, …, 399

Common difference = 111 − 105 = 6

To get the nth natural number from the 1st natural number, we must add the common difference (n 1) times to the 1st natural number.

399 = 105 + (n − 1) × 6

399 − 105 = 6n − 6

6n − 6 = 294

6n = 300 

n = 50

Thus, there are 50 terms between 100 and 400 which leave the remainder 3 when divided by 6.



Page No 20:

Question 1:

The first term and common difference of some arithmetic sequence are given below. Write each of them in the form xn = an + b; also write down the first three terms of each:

(i) first term −2, common difference 5

(ii) first term 2, common difference −5

(iii) first term 1, common difference

Answer:

(i)

First term of the arithmetic sequence = a + b = 2 … (1)

Common difference = a = 5

Putting the value of a in equation (1):

5 + b = 2

b = −2 − 5 = 7

Now, second term = 5(2) − 7 = 10 − 7 = 3

Third term = 5(3) − 7 = 15 − 7 = 8

Therefore, the first three terms of the given arithmetic sequence are 2, 3 and 8.


 

(ii)

First term of the arithmetic sequence = a + b = 2 … (1)

Common difference = a = 5

Putting the value of a in equation (1):

5 + b = 2

b = 2 + 5 = 7

Now, second term = 5(2) + 7 = 10 + 7 = 3

Third term = 5(3) + 7 = 15 + 7 = 8

Therefore, the first three terms of the given arithmetic sequence are 2, 3 and 8.


 

(iii)

First term of the arithmetic sequence = a + b = 1 … (1)

Common difference = a =

Putting the value of a in equation (1):

+ b = 2

Now, second term =

Third term =

Therefore, the first three terms of the given arithmetic sequence are 2, and 3.



Page No 21:

Question 1:

The algebraic forms of some sequence are given below. Check whether each of them is an arithmetic sequence or not; also find out the first term and the common difference of the arithmetic sequence

(i) xn = 4 − 3n

(ii) xn = n2 + 2

(iii)

(iv)

(v) xn = (n + 1)2 − (n − 1)2

Answer:

(i)

We know that every sequence of the form is an arithmetic sequence.

Given sequence is , which can also be written as

On comparing with , we get that the given sequence is an arithmetic sequence.

Common difference = a = 3

First term = a + b = 3 + 4 = 1


 

(ii)

We know that every sequence of the form is an arithmetic sequence.

Given sequence is

On comparing with , we get that the given sequence is not an arithmetic sequence.


 

(iii)

We know that every sequence of the form is an arithmetic sequence.

Given sequence is , which can also be written as

On comparing with , we get that the given sequence is an arithmetic sequence.

Common difference = a =

First term = a + b =


 

(iv)

We know that every sequence of the form is an arithmetic sequence.

Given sequence is , which can also be written as

On comparing with , we get that the given sequence is not an arithmetic sequence.


 

(v)

We know that every sequence of the form is an arithmetic sequence.

Given sequence is

Now, this sequence can also be written as

On comparing with , we get that the given sequence is an arithmetic sequence.

Common difference = a = 4

First term = a + b = 4 + 0 = 4

Page No 21:

Question 2:

Multiply each odd number by 2 and add 1. If these numbers are written in order, does it form and arithmetic sequence? Write the algebraic form of this sequence. Now consider the odd numbers not in this sequence. Do they form and arithmetic sequence? What is its algebraic form?

Answer:

The odd numbers are 1, 3, 5, …

When we multiply each odd number by 2 and add 1, then the sequence becomes:

(1 × 2) + 1, (3 × 2) + 1, (5 × 2) + 1, …

Or 3, 7, 11, …

Difference between second and first term = 7 − 3 = 4

Difference between third and second term = 11 − 7 = 4

Since the difference between the consecutive terms is same, the sequence is an arithmetic sequence with common difference = a = 4.

First term = a + b = 3

4 + b = 3

b = 3 − 4 = 1

Thus, the algebraic form of the given sequence is 4n − 1. 

The odd numbers not in the above sequence are 1, 5, 9, 13, …

Difference between second and first term = 5 − 1 = 4

Difference between third and second term = 9 − 5 = 4

Since the difference between the consecutive terms is same, the sequence is an arithmetic sequence with common difference = a = 4.

First term = a + b = 1

4 + b = 1

b = 1 − 4 = 3

Thus, the algebraic form of the given sequence is 4n − 3. 

Page No 21:

Question 3:

Consider the arithmetic sequence with first termand common difference. Is 1 a term of this sequence? What about 2?

Write the algebraic form of this sequence. Prove that all natural numbers occur in this sequence.

Answer:

First term of the given sequence = a + b =

Common difference = a =

Let 1 be the nth term of the given sequence.

Thus, 1 is the 3rd term of this sequence.

Now, let 2 be the mth term of the given sequence. 

Thus, 2 is the 7th term of this sequence.

Therefore, 1 and 2 are the 3rd and the 7th terms of this sequence.

The algebraic form of the sequence is

Now, let 3 be the pth term of the given sequence. 

Thus, 3 is the 11th term of this sequence.

3rd term of the given sequence is 1, 7th term is 2, 11th term is 3 and so on.

Thus, the new sequence obtained is 3, 7, 11, …

Here, first term = a + b = 3

Common difference = a = 7 − 3 = 4

4 + b = 3

b = 3 − 4 = 1 

The algebraic form of the sequence is, where q is a natural number.

xq will surely satisfy as xq is the general form of the nth term of this sequence which gives a natural number as the result. 

As q is a natural number, so all the natural numbers occur in the sequence

Page No 21:

Question 4:

Consider the arithmetic sequence with first term and common difference. Is 1 a term of this sequence? What about 2?

Write the algebraic form of this sequence. Prove the no natural number occurs is this sequence.

Answer:

First term of an arithmetic sequence = a + b =

Common difference = a =

Let 1 be the nth term of the given sequence. 

As n must be a natural number, 1 is not a term of this sequence.

Now, let 2 be the mth term of the given sequence.

As n must be a natural number, 2 is not a term of this sequence.

Algebraic form of the sequence is

Let p, a natural number be the nth term of the given sequence.

Therefore, for any value of p, n cannot be a natural number.

Thus, no natural number occurs in this sequence.

Page No 21:

Question 5:

In an arithmetic sequence, the ratio of the first term to the second is 2 : 3. What is the ratio of the third term to the fifth term?

Answer:

Let the first and the second term of the arithmetic sequence be 2x and 3x respectively.

a + b = 2x … (1)

Also, 2a + b = 3x

a + a + b = 3x

a + 2x = 3x (From equation (1))

a = 3x − 2x = x

Putting the value of a in equation (1):

x + b = 2x

b = 2xx = x

Third term of the given arithmetic sequence = 3a + b = 3x + x = 4x

Fifth term of the given arithmetic sequence = 5a + b = 5x + x = 6x

Ratio of the third term to the fifth term =



Page No 25:

Question 1:

The first term of an arithmetic sequence is 5 and the common difference is 2. Find the sum of its first 25 terms.

Answer:

First term of an arithmetic sequence = a + b = 5

Common difference = a = 2

2 + b = 5

b = 5 − 2 = 3

Therefore, nth term of the given arithmetic sequence is

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 25 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 25 terms =

Page No 25:

Question 2:

Find and algebraic expression to compute the sum of the first n terms of the arithmetic sequence with first term f and common difference d.

Answer:

First term of an arithmetic sequence = f

Common difference = d

Therefore, the arithmetic sequence is f, f + d, f + 2d, f + 3d, …

This can also be written as f + (1 − 1)d, f + (2 − 1)d, f + (3 − 1)d, f + (4 − 1)d, …

Thus, its nth term,

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first n terms = … (1)

Here,

Putting the values in equation (1):

Sum of first n terms =

Page No 25:

Question 3:

If 52 × 54 × 56 × ……× 52n = (0.04)−28 what is n?

Answer:

The given equation is

5(2 + 4 + 6 + … + 2n) = (0.04)28 … (1)

2 + 4 + 6 + … + 2n

= 2(1 + 2 + 3 + … + n)

From (1) and (2), we get:

n2 + n − 56 = 0

n2 + 8n − 7n − 56 = 0 

n(n + 8) − 7(n + 8) = 0

(n + 8)(n − 7) = 0

n = 8 or 7

Page No 25:

Question 4:

Find the sum of all three digit numbers which are multiples of 9.

Answer:

First three-digit number which is a multiple of 9 = 108

Second three-digit number which is a multiple of 9 = 117

Third three-digit number which is a multiple of 9 = 126

nth three digit number which is a multiple of 9 = 999

So, the arithmetic sequence is 108, 117, 126, …, 999

Common difference = a = 117 − 108 = 9

First term of the arithmetic sequence = a + b = 108

9 + b = 108

b = 108 − 9 = 99

Therefore,

999 = 9n + 99

999 − 99 = 9n

900 = 9n

n = 100

Therefore, there are 100 three-digit numbers, which are the multiples of 9.

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 100 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 100 terms =

Page No 25:

Question 5:

Prove that the sum of the first five terms of any arithmetic sequence is five times the middle number. What about the sum of the first seven terms? Can you formulate a general principle from these examples?

Answer:

(1) 

Let the first five terms of an arithmetic sequence be a + b, 2a + b, 3a + b, 4a + b and 5a + b.

Here, the first term = a + b

Common difference = a 

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 5 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 5 terms =

= Five times the middle number, i.e., the third number, 3a + b

Therefore, the sum of the first five terms is five times the middle term.


 

(2) 

Let the first seven terms of an arithmetic sequence be a + b, 2a + b, 3a + b, 4a + b, 5a + b, 6a + b and 7a + b.

Here, first term = a + b

Common difference = a 

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 7 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 7 terms

= Seven times the middle number, i.e., the fourth number, 4a + b

Therefore, the sum of the first seven terms is seven times the middle term.

General principle says that the sum of first n terms of an arithmetic sequence, where n is an odd number is n times the middle number.



Page No 26:

Question 1:

The sum of the first n terms of an arithmetic sequence is 2n2 + 3n. find the algebraic form of this sequence.

Answer:

Sum of first n terms of an arithmetic sequence = 2n2 + 3n

First term of the arithmetic sequence = x1 = 2(1)2 + 3(1) 

= 2 + 3 

= 5

Sum of first two terms = x1 + x2 = 2(2)2 + 3(2) 

= 8 + 6 

= 14

5 + x2 = 14

x2 = 14 − 5 = 9

Sum of first three terms = x1 + x2 + x3 = 2(3)2 + 3(3) 

= 18 + 9 

= 27

5 + 9 + x3 = 27

x3 = 27 − 14 = 13

Thus, the arithmetic sequence is 5, 9, 13, …

Here, a + b = 5 and common difference = a = 9 − 5 = 4

4 + b = 5

b = 5 − 4 = 1

Therefore,

Thus, the algebraic form of the given sequence is 4n + 1, where n is a natural number.

Page No 26:

Question 2:

Consider the arithmetic sequences 3, 5, 7, ….. and 4, 6, 8, ….. How much more is the sum of the first 25 terms of the second than the sum of the first 25 terms of the first?

Answer:

Consider the first arithmetic sequence 3, 5, 7, …

First term = a + b = 3

Common difference = a = 5 − 3 = 2

2 + b = 3

b = 3 − 2 = 1

Thus, the nth term of the first arithmetic sequence is:

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 25 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 25 terms of the first arithmetic sequence

Now, consider the second arithmetic sequence 4, 6, 8, …

First term = a + b = 4

Common difference = a = 6 − 4 = 2

2 + b = 4

b = 4 − 2 = 2

Thus, the nth term of the first arithmetic sequence is:

Sum of its first 25 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 25 terms of the second arithmetic sequence =

Required difference = 700 − 675 = 25

Thus, the sum of first 25 terms of the second arithmetic sequence is 25 more than the sum of first 25 terms of the first arithmetic sequence.

Page No 26:

Question 4:

51 + 52 + 53 + ……. + 70

Answer:

The given series is 51 + 52 + 53 + … + 70.

Here, first term = a + b = 51

Common difference = a = 52 − 51 = 1

1 + b = 51

b = 50

Therefore,

70 = n + 50

n = 70 − 50 = 20

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 20 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 20 terms =

Page No 26:

Question 5:

Answer:

The given series is

This can also be written as

Consider the series 1 + 3 + 5 + … + 25.

Here, first term = a + b = 1

Common difference = a = 3 − 1 = 2

2 + b = 1

b = 1 − 2 = 1

Therefore,

25 = 2n 1

2n = 26

n = 13

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 13 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 13 terms =

1 + 3 + 5 + … + 25 = 169

Thus,

Page No 26:

Question 6:

Answer:

The given series is

This can also be written as

We know that the sum of all natural numbers starting from one to a specified natural number is half the product of the number and the next natural number.

1 + 2 + 3 + 4 + 5 + … + 25 =

1 + 2 + 3 + 4 + 5 + … + 25 = 325

Thus,

Page No 26:

Question 3:

How much more is the sum of the natural numbers from 21 to 40 than the sum of the natural numbers from 1 to 20?

Answer:

Natural numbers from 21 to 40 = 21, 22, 23, … , 40

Sum of these numbers = 21 + 22 + 23 + … + 40 

Here, first term = a + b = 21

Common difference = a = 22 − 21 = 1

1 + b = 21

b = 20

Therefore,

40 = n + 20

n = 40 − 20 = 20

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

Sum of its first 20 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 20 terms =

Natural numbers from 1 to 20 = 1, 2, 3, … , 20

Sum of these numbers = 1 + 2 + 3 + … + 20

Here, first term = a + b = 1

Common difference = a = 2 − 1 = 1

1 + b = 1

b = 0

Therefore,

n = 20

Sum of its first 20 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 20 terms =

Required difference = 610 − 210 = 400

Thus, the sum of natural numbers from 21 to 40 is 400 more than the sum of natural numbers from 1 to 20.



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