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

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions

**Exercise 11B**

**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)

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.