# RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 11 Arithmetic Progressions Ex 11B Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 11 Arithmetic Progressions Ex 11B Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B Maths book pdf download. Now you will get step by step solution to each question.

### RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11BPDF

Question 1.
Solution:
(3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
(4k – 6) – (3k – 2) = (k + 2) – (4k – 6)
⇒ 2(4k – 6) = (k + 2) + (3k – 2)
⇒ 8k – 12 = 4k + 0
⇒ 8k – 4k = 0 + 12
⇒ 4k = 12
k = 3

Question 2.
Solution:
(5x + 2), (4x – 1) and (x + 2) are in AP.
(4x – 1) – (5x + 2) = (x + 2) – (4x – 1)
⇒ 2(4x – 1) = (x + 2) + (5x + 2)
⇒ 8x – 2 = 6x + 2 + 2
⇒ 8x – 2 = 6x + 4
⇒ 8x – 6x = 4 + 2
⇒ 2x = 6
x = 3

Question 3.
Solution:
(3y – 1), (3y + 5) and (5y + 1) are the three consecutive terms of an AP.
(3y + 5) – (3y – 1) – (5y + 1) – (3y + 5)
⇒ 2(3y + 5) = 5y + 1 + 3y – 1
⇒ 6y + 10 = 8y
⇒ 8y – 6y = 10
⇒ 2y = 10
⇒ y = 5
y = 5

Question 4.
Solution:
(x + 2), 2x, (2x + 3) are three consecutive terms of an AP.
2x – (x + 2) = (2x + 3) – 2x
⇒ 2x – x – 2 = 2x + 3 – 2x
⇒ x – 2 = 3
⇒ x = 2 + 3 = 5
x = 5

Question 5.
Solution:
(a – b)², (a² + b²) and (a + b)² will be in AP.
If (a² + b²) – (a – b)² = (a + b)² – (a² + b²)
If (a² + b²) – (a² + b² – 2ab) = a² + b² + 2ab – a² – b²
2ab = 2ab which is true.
Hence proved.

Question 6.
Solution:
Let the three numbers in AP be
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
and (a – d) x a x (a + d) = 80
a(a² – d²) = 80
⇒ 5(5² – d²) = 80
⇒ 25 – d² = 16
⇒ d² = 25 – 16 = 9 = (±3)²
d = ±3
Now, a = 5, d = +3
Numbers are 5 – 3 = 2
5 and 5 + 3 = 8
= (2, 5, 8) or (8, 5, 2)

Question 7.
Solution:
Let the three numbers in AP be a – d, a and a + d Question 8.
Solution:
Sum of three numbers = 24
Let the three numbers in AP be a – d, a, a + d Question 9.
Solution:
Let three consecutive in AP be a – d, a, a + d
a – d + a + a + d = 21
⇒ 3a = 21 Question 10.
Solution:
Sum of angles of a quadrilateral = 360°
Let d= 10
The first number be a, then the four numbers will be
a, a + 10, a + 20, a + 30
a + a + 10 + a + 20 + a + 30 = 360
4a + 60 = 360
4a = 360 – 60 = 300
Angles will be 75°, 85°, 95°, 105°

Question 11.
Solution:
Let the four numbers in AP be a – 3d, a – d, a + d, a + 3d, then
a – 3d + a – d + a + d + a + 3d = 28 Question 12.
Solution:
Let the four parts of 32 be a – 3d, a – d, a + d, a + 3d Question 13.
Solution:
Let the three terms be a – d, a, a + d
a – d + a + a + d = 48
⇒ 3a = 48
⇒ a = 16
and (a – d) x a = (a + d) + 12
⇒ a(a – d) = 4 (a + d) + 12
⇒ 16 (16 – d) = 4(16 + d) + 12
⇒ 256 – 16d = 64 + 4d + 12 = 4d + 76
⇒ 256 – 76 = 4d + 16d
⇒ 180 = 20d
⇒ d = 9
Numbers are (16 – 9, 16), (16 + 9) or (7, 16, 25)

All Chapter RS Aggarwal Solutions For Class 10 Maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class10

All Subject NCERT Solutions For Class 10

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.