# RS Aggarwal Solutions Chapter 11 Arithmetic Progression Exercise 11B Class 10 Maths

 Chapter Name RS Aggarwal Chapter 11 Arithmetic Progression Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 11AExercise 11CExercise 11DExercise 11E Related Study NCERT Solutions for Class 10 Maths

### Exercise 11B Solutions

1. Determine k so that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.

Solution

It is given that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.

∴ (4k – 6) – (3k – 2) = (k + 2) – (4k – 6)

⇒ 4k – 6 – 3k + 2 = k + 2 – 4k + 6

⇒ k – 4 = -3k + 8

⇒ k + 3k = 8 + 4

⇒ 4k = 12

⇒ k = 3

Hence, the value of k is 3.

2. Find the value of x for which the numbers (5x + 2), (4x – 1) and (x + 2) are in AP.

Solution

It is given that (5x + 2), (4x – 1) and (x + 2) are in AP.

∴ (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)

⇒ 4x – 1 – 5x – 2 = x + 2 – 4x + 1

⇒ - x – 3 = - 3x + 3

⇒ 2x = 6

⇒ x = 3

Hence, the value of x is 3.

3. If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.

Solution

It is given that (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.

∴ (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)

⇒ 3y + 5 – 3y + 1 = 5y + 1 – 3y – 5

⇒ 6 = 2y – 4

⇒ 2y = 6 + 4 = 10

⇒ y = 5

Hence, the value of y is 5.

4. Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.

Solution

Since (x + 2), 2x and (2x + 3) are in AP, we have

2x – (x + 2) = (2x + 3) – 2x

⇒ x – 2 = 3

⇒ x = 5

∴ x = 5

5. Show that (a – b)2, (a2 + b2) and (a2 + b2) are in AP.

Solution

Now,

(a2 + b2) – (a – b)2 = a2 + b2 – (a2 – 2ab + b2) = a2 + b2 – a2 + 2ab – b2 = 2ab

(a + b)2 – (a2 + b2) = a2 + 2ab + b2 – a2 – b2 = 2ab

So, (a2 + b2) – (a – b)2 = (a + b)2 – (a2 + b2) = 2ab (Constant)

Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.

6. Find the three numbers in AP whose sum is 15 and product is 80.

Solution

Let the required numbers be (a – d), a and (a + d).

Then (a – d) + a + (a + d) = 15

⇒ 3a = 15

⇒ a = 5

Also, (a – d).a.(a + d) = 80

⇒ a(a2 – d2) = 80

⇒ 5(25 – d2) = 80

⇒ d2 = (25 – 16) = 9

⇒ d = ± 3

Thus, a = 5 and d = ±3

Hence, the required numbers are (2, 5 and 8) or (8, 5 and 2).

7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers.

Solution

Let the required numbers be (a – d), a and (a + d).

Then (a – d) + a + (a + d) = 3

⇒ 3a = 3

⇒ a = 1

Also, (a – d), a.(a + d) = - 35

⇒ a(a2 – d2) = - 35

⇒ 1.(1 – d2) = - 35

⇒ d2 = 36

⇒ d = ± 6

Thus, a = 1 and d = ± 6

Hence, the required numbers are (-5, 1 and 7) or (7, 1 and -5).

8. Divide 24 in three parts such that they are in AP and their product is 440.

Solution

Let the required parts of 24 be (a – d), a and (a + d) such that they are in AP.

Then (a – d) + a + (a + d) = 24

⇒ 3a = 24

⇒ a = 8

Also, (a – d).a.(a + d) = 440

⇒ a(a2 – d2) = 440

⇒ 8(64 – d2) = 440

⇒ d2 = 64 – 55 = 9

⇒ d = ±3

Thus, a = 8 and d = ±3

Hence, the required parts of 24 are (5, 8, 11) or (11, 8, 5).

9. The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.

Solution

Let the required terms be (a – d), a and (a + d).

Then (a – d) + a + (a + d) = 21

⇒ 3a = 21

⇒ a = 7

Also, (a – d)2 + a2 + (a + d)2 = 165

⇒ 3a2 + 2d2 = 165

⇒ (3 × 49 + 2d2) = 165

⇒ 2d2 = 165 – 147 = 18

⇒ d2 = 9

⇒ d = ±3

Thus, a = 7 and d = ±3

Hence, the required terms are (4, 7, 10) or (10, 7, 4).

10. The angles of quadrilateral are in whose AP common difference is 10°. Find the angles.

Solution

Let the required angles be (a – 15)°, (a – 5)°, (a + 5)° and (a + 15)°, as the common difference is 10 (given).

Then (a – 15)° + (a – 5)° + (a + 5)° + (a + 15)° = 360°

⇒ 4a = 360

⇒ a = 90

Hence, the required angles of a quadrilateral are (90 – 15)°, (90 – 5)°, (90 + 5)° and (90 + 15)°; or 75°, 85°, 95°, and 105°.

11. Find four numbers in AP whose sum is 8 and the sum of whose squares is 216.

Solution

(4, 6, 8, 10) or (10, 8, 6, 4)

12 Divide 32 into four parts which are the four terms of an AP such that the product of the first and fourth terms is to product of the second and the third terms as 7 : 15.

Solution

Let the four parts in AP be (a – 3d), (a – d), (a + d) and (a + 3d). Then,

(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32

⇒ a = 8 ...(1)

Also,

(a – 3d)(a + 3d): (a – d)(a + d) = 7 : 15

⇒ (8 – 3d)(8 + 3d)/(8 – d)(8 + d) = 7/15 [From (1)]

⇒ (64 – 9d2)/(64 – d2) = 7/15

⇒ 15(64 – 9d2) = 7(64 – d2)

⇒ 960 – 135d2 = 448 – 7d2

⇒ 135d2 – 7d2 = 960 – 448

⇒ 128d2 = 512

⇒ d2 = 4

⇒ d = ±

When a = 8 and d = 2,

a – 3d = 8 – 3 × 2 = 8 – 6 = 2

a – d = 8 – 2 = 6

a + d = 8 + 2 = 10

a + 3d = 8 + 3 × 2 = 8 + 6 = 14

When a = 8 and d = - 2,

a – 3d = 8 – 3 × (-2) = 8 + 6 = 14

a – d = 8 – (-2) = 8 + 2 = 10

a + d = 8 – 2 = 6

a + 3d = 8 + 3 × (-2) = 8 – 6 = 2

Hence, the four parts are 2, 6, 10 and 14.

13. The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times and third term by 12. Find the AP.

Solution

Let the first three terms of the AP be (a – d), a and (a + d). Then,

(a – d) + a + (a + d) = 48

⇒ 3a = 48

⇒ a = 16

Now,

(a – d) × a = 4(a + d) + 12 (Given)

⇒ (16 – d) × 16 = 4(16 + d) + 12

⇒ 256 – 16d = 64 + 4d + 12

⇒ 16d + 4d = 256 – 76

⇒ 20d = 180

⇒ d = 9

When a = 16 and d = 9,

a – d = 16 – 9 = 7

a + d = 16 + 9 = 25

Hence, the first three terms of the AP are 7, 16 and 25.