R S Aggarwal Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5D

R S Aggarwal Solutions for Class 10 Maths exercise 5d are available here. Students can get detailed step-by-step explanations to all the questions listed under exercise 5d of Arithmetic Progression of the Class 10 R S Aggarwal Textbook. Download pdf of Class 10 Maths Chapter 5 R S Aggarwal Solutions and accelerate your knowledge.

Download PDF of R S Aggarwal Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5D

RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D
RS Aggarwal Sol class 10 Maths Chapter 5D

 

Access Answers of Maths R S Aggarwal Class 10 Chapter 5 Arithmetic Progression Exercise 5D Page number 289

Access other exercise solutions of Class 10 Maths Chapter 5 Arithmetic Progression

Exercise 5A Solutions: 30 Questions (Short Answers)

Exercise 5B Solutions: 16 Questions (Short Answers)

Exercise 5C Solutions: 15 Questions (Short Answers)

Exercise 5D Page No: 289

Question 1: The first three terms of an AP are respectively (3y – 1), (3y + 5) and (5y + 1), find the value of y.

Solution:

Given: (3y – 1), (3y + 5) and (5y+ 1) are in AP

So, (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)

2 (3y + 5) = (5y + 1) + (3y – 1)

6y + 10 = 8y

8y – 6y = 10

2y = 10

Or y = 5

The value of y is 2.

Question 2: If k, (2k – 1) and (2k + 1) are the three successive terms of an AP, find the value of k.

Solution:

Given: k, (2k – 1) and (2k + 1) are the three successive terms of an AP.

So, (2k – 1) – k = (2k + 1) – (2k – 1)

2 (2k – 1) = 2k + 1 + k

4k – 2 = 3k + 1

4k – 3k = 1 + 2

or k = 3

The value of k is 3.

Question 3: If 18, a, (b – 3) are in AP, then find the value of (2a – b).

Solution:

Given: 18, a, (b – 3) are in AP

a – 18 = b – 3 – a

a + a – b = -3 + 18

2a – b = 15. Answer!

Question 4: If the numbers a, 9, b, 25 form an AP, find a and b.

Solution:

Given: a, 9, b, 25 are in AP.

So, 9 – a = b – 9 = 25 – b

b – 9 = 25 – b

b + b = 22 + 9 = 34

or b = 17

And,

a – b = a – 9

9 + 9 = a + b

a + b = 18

a + 17 = 18

or a = 1

Answer: a = 18, b= 17

Question 5: If the numbers (2n – 1), (3n + 2) and (6n – 1) are in AP, find the value of n and the numbers.

Solution:

Given: (2n – 1), (3n + 2) and (6n – 1) are in AP

So, (3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)

(3n + 2) + (3n + 2) = 6n – 1 + 2n – 1

6n + 4 = 8n – 2

8n – 6n = 4 + 2

Or n = 3

Numbers are:

2 x 3 – 1 = 5

3 x 3 + 2 = 11

6 x 3 – 1 = 17

Answer: (5, 11, 17) are required numbers.

Question 6: How many three-digit natural numbers are divisible by 7?

Solution:

3-digit natural numbers: 100, 101,…….. 990 and

3-digit natural numbers divisible by 7: 105, 112, 119, 126, …, 994\

Here, a = 105, d= 7, l = 994

a_n = (l) = a + (n – 1) d

994 = 105 + (n – 1) x 7

994 – 105 = (n – 1) 7

n – 1 = 127

n = 128

Answer: There are 128 required numbers.

Question 7: How many three-digit natural numbers are divisible by 9?

Solution:

3-digit numbers: 100, 101,…….,999

3-digit numbers divisible by 9 : 108, 117, 126, 135, …, 999

Here, a = 108, d= 9, l = 999

a_n (l) = a + (n – 1) d

999 = 108 + (n – 1) x 9

(n – 1) x 9 = 999 – 108 = 891

n – 1 = 99

n = 100

Question 8: If the sum of first m terms of an AP is (2m² + 3m) then what is its second term?

Solution:

Sum of first m terms of an AP = 2m² + 3m (given)

S_m = 2m² + 3m

Sum of one term = S_1 = 2(1)² + 3 x 1 = 2 + 3 = 5 = first term

Sum of first two terms = S_2 = 2(2)² + 3 x 2 = 8 + 6=14

Sum of first three terms = S_3 = 2(3)² + 3 x 3 = 18 + 9 = 27

Now,

Second term = a_2 = S_2 – S_1 = 14 – 5 = 9

Question 9: What is the sum of first n terms of the AP, a, 3a, 5a,………….

Solution:

AP is a, 3a, 5a,…..

Here, a = a, d = 2a

Sum = S_n = n/2 [2a + (n-1)d]

= n/2[2a + 2an – 2a]

=an^2

Question 10.: What is the 5th term from the end of the AP 2, 7, 12, ……… 47?

Solution:

Given AP is 2, 7, 12, 17, …… 47

Here, a = 2, d = 7 – 2 = 5, l = 47

nth term from the end = l – (n – 1 )d

5th term from the end = 47 – (5 – 1) 5 = 47 – 4 x 5 = 27

Question 11: If an denotes the nth term of the AP 2, 7, 12, 17, …, find the value of (a_30 – a_20).

Solution:

Given AP is 2, 7, 12, 17,……..

Here, a = 2, d = 7 – 2 = 5

Now,

a_n = a + (n – 1) d = 2 + (n – 1) 5 = 5n – 3

a_30 = 2 + (30 – 1) 5 = 2 + 145 = 147 and

a_20 = 2 + (20 – 1) 5 = 2 + 95 = 97

a_30 – a_20 = 147 – 97 = 50

Question 12: The nth term of an AP is (3n + 5). Find its common difference.

Solution:

Nth term = a_n = 3n + 5 (given)

a_(n-1) = 3 (n – 1) + 5 = 3n + 2

Common difference = d = a_n – a_(n-1)

= (3n + 5) – (3n + 2)

= 3n + 5 – 3n – 2

= 3

Therefore, common difference is 3.

Question 13: The nth term of an AP is (7 – 4n). Find its common difference.

Solution:

Nth term = a_n = 7 – 4n

a_(n-1) = 7 – 4(n – 1) = 11 – 4n

Common difference = d = a_n – a_(n-1)

= (7 – 4n) – (11 – 4n)

= 7 – 4n – 11 + 4n

= -4

Therefore, common difference is -4.

Question 14: Write the next term of the AP √8, √18, √32, ……..

Solution:

Given AP is √8, √18, √32,……..

Above AP can be written as:

2√2 , 3√2, 4 √2, ……

Here a = 2√2 and d = √2

Next term = 4√2 + √2 = 5√2 = √50

Question 15: Write the next term of the AP √2, √8, √18,….

Solution:

Given AP is √2, √8, √18,….

Can be written as:

√2, 2√2, 3√2,…..

First term = √2

Common difference = 2√2 – √2 = √2

Next term = 3√2 + √2 = 4√2 = √32

Question 16: Which term of the AP 21, 18, 15,…. is zero?

Solution:

Given AP is 21, 18, 15,….

First term = a = 21

Common difference = d = 18-21 = -3

Last term = l = 0

l = a + (n – 1) d

0 = 21 + (n – 1)(-3)

0 = 21 – 3n + 3

24 – 3n = 0

Or n = 8

Answer: Zero is the 8th term.

Question 17: Find the sum of first n natural numbers.

Solution:

First n natural numbers: 1, 2, 3, 4, 5, …, n

Here, a = 1, d = 1

Sum = S_n = n/2 [2a + (n-1)d]

= n/2 [2(1) + (n-1)(1)]

= n(n+1)/2

Question 18: Find the sum of first n even natural numbers.

Solution:

First n even natural numbers: 2, 4, 6, 8, 10,….,n

Here, a = 2, d = 4 – 2 = 2

Sum = S_n = n/2 [2a + (n-1)d]

= n/2 [2(2) + (n-1)(2)]

= n(n+1)

Question 19: The first term of an AP is p and its common difference is q. Find its 10th term.

Solution:

Given:

First term =a = p and

Common difference = d =q

Now,

a_10 = a + (n – 1) d

= p + (10 – 1)q

= (p + 9q)

Question 20: If 4/5, a, 2 are in AP, find the value of a.

Solution:

AP terms: 4/5, a, 2 (given)

Then,

a – 4/5 = 2 – a

a = 7/5

Question 21: If (2p + 1), 13, (5p – 3) are in AP, find the value of p.

Solution:

Given, 2p + 1, 13, 5p – 3 are in AP

Then,

13 – (2p + 1) = (5p – 3) – 13

13 – 2p – 1 = 5p – 3 – 13

12 – 2p = 5p – 16

p = 4

The value of p is 4.

Question 22: If (2p – 1), 7, 3p are in AP, find the value of p.

Solution:

Given, (2p – 1), 7, 3p are in AP

Then,

7 – (2p – 1) = 3p – 7

7 – 2p + 1 = 3p – 7

5p = 15

p = 3

The value of p is 3.

Question 23: If the sum of first p terms of an AP is (ap² + bp), find its common difference.

Solution:

Sum of first p terms = S_p = (ap² + bp)

Sum of one term = S_1 = a(1)² + b(1) = a+b = first term

Sum of first two terms = S_2 = a(2)² + b x 2 = 4a + 2b

We know that, second term = a_2 = S_2 – S_1

= (4a + 2b) – (a + b)

= 3a + b

Now, d = a_2 – a_1

= 3a + b – (a+b)

= 2a

Answer: Common difference is 2a.

Question 24: If the sum of first n terms is (3n² + 5n), find its common difference.

Solution:

Sum of first n terms = S_n = (3n² + 5n)

Sum of one term = S_1 = 3(1)² + 5(1) = 8 = first term

Sum of first two terms = S_2 = 3(2)² + 5(2) = 22

We know that, second term = a_2 = S_2 – S_1

=22 – 8

= 14

=> a_2 = 14

Now, d = a_2 – a_1

= 14 – 8 = 6

Answer: Common difference is 6.

Question 25: Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Solution:

Let a be the first term and d be the common difference.

Given:

4th term = a_4 = 9

Sum of 6th and 13th terms = a_6 + a_13 = 40

Now,

a_4 = a + (4-1)d

9 = a + 3d

a = 9 – 3d ….(1)

And

a_6 + a_13 = 40

a + 5d + a + 12d = 40

2a +17d = 40

2(9 – 3d) + 17d = 40 (using (1))

d = 2

From (1): a = 9 – 6 = 3

Required AP = 3,5,7,9,…..

Question 26: What is the common difference of an AP in which a_27 – a_7 = 84?

Solution:

Given: a_27 – a_7 = 84

[a + 26d] – [a + 6d] = 84

20d = 84

d = 4.2

Question 27: If 1 + 4 + 7 + 10 + … + x = 287, find the value of x.

Solution:

Given: 1 + 4 + 7 + 10 + … + x = 287

Here a = 1, d = 3 and S_n = 287

Sum = S_n = n/2 [2a + (n-1)d]

287 = n/2 [2 + (n-1)3]

574 = 3n^2 – n

Which is a quadratic equation.

Solve 3n^2 – n – 574 = 0

3n^2 – 42n + 41n – 574 = 0

3n(n – 14) + 41(n-14) = 0

(3n + 41)(n-14) = 0

Either (3n + 41) = 0 or (n-14) = 0

n = -41/3 or n = 14

Since number of terms cannot be negative, so result is n = 14.

=> Total number of terms in AP are 14.

Which shows, x = a_14

or x = a + 13d

or x = 1 + 39

or x = 40

The value of x is 40.

R S Aggarwal Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5D

Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5D is based on the following topics:

  • General term of an AP: a_n = a + (n-1)d
  • nth term from the end of an AP
  • Arithmetic mean between two numbers
  • Sum of n terms of an AP