# RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progression Exercise 9.5

**1. ****Find the value of x for which (8x + 4), (6x - 2) and (2x + 7) are in A.P.**

**Solution**

Then, 2(6x - 2) = 8x + 4 + 2x + 7

12x - 4 = 10 + 11

⇒ 2x = 15

⇒ x = 15/2

**2. If x + 1, 3x and 4x + 2 are in A.P., find the value of x.**

**Solution**

If a, b, c are the given numbers are in AP then, 2b = a + c

Then,

⇒ 6x = 5x + 3

⇒ x = 3

**8. Show that (a – b)**

^{2}, (a^{2}+ b^{2}) and (a + b)^{2}are in A.P.**Solution**

^{2}, (a

^{2}+ b

^{2}) and (a + b)

^{2}are in A.P.

If they are in AP. Then they have to satisfy the condition

2b = a + c

**4. The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.**

**Solution**

Let the three terms of an A.P. be a – d, a, a + d

Sum of three terms = 21

⇒ a – d + a + a + d = 21

⇒ 3a = 21

⇒ a = 7

and product of the first and 3rd = 2nd term + 6

⇒ (a – d) (a + d) = a + 6

a² – d² = a + 6

⇒ (7)² – d² = 7 + 6

⇒ 49 – d² = 13

⇒ d² = 49 – 13 = 36

⇒ d² = (6)²

⇒ d = 6

Terms are 7-6, 7, 7+6

⇒ 1, 7, 13

**5. Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.**

**Solution**

Let the three numbers of an A.P. be a – d, a, a + d

According to the conditions,

Sum of these numbers = 27

a – d + a + a + d = 27

⇒ 3a = 27

⇒ d = 3

Numbers are a-d, a, a+d or, 9-3, 9, 9+3

⇒ 6, 9, 12

**6. Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.**

**Solution**

Let the four terms of an A.P. be (a – 3d), (a – d), (a + d) and (a + 3d)

Now according to the condition,

Sum of these terms = 50

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50

⇒ a – 3d + a – d + a + d + a – 3d= 50

⇒ 4a = 50

⇒ a = 25/2

and greatest number = 4 x least number

⇒ a + 3d = 4 (a – 3d)

⇒ a + 3d = 4a – 12d

⇒ 4a – a = 3d + 12d

The terms are 5, 10, 15, 20.

**7. The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.**

**Solution**

Let the three numbers in A.P. be a-d, a, a+d.

a - d + a + a + d = 12

⇒ 3a = 12

⇒ a = 12/3 = 4

Numbers will be 4-2, 4, 4+2 or 2, 4, 6

or, 4+2, 4, 4-2 or 6, 4, 2

**8. Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.**

**Solution**

Let four numbers of A.P. be (a-3d), (A-d), (a+d), (a+3d)

According to question,

Required numbers are:

a - 3d = 14 - 3×2 = 8

a - d = 14 - 2 = 12

a + d = 14 + 2 = 16

a + 3d = 14 + 3×2 = 20

The A.P. is 8, 12, 16, 20.

**9. The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.**

**Solution**

Let the four angles of a quadrilateral which are in A.P., be

a – 3d, a – d, a + d, a + 3d

Common difference = 10°

Now sum of angles of a quadrilateral = 360°

a – 3d + a – d + a + d + a + 3d = 360°

⇒ 4a = 360°

⇒ a = 90°

and common difference = (a – d) – (a – 3d) = a – d – a + 3d = 2d

2d = 10°

⇒ d = 5°

Angles will be

a – 3d = 90° – 3×5° = 90° – 15° = 75°

a – d= 90° – 5° = 85°

a + d = 90° + 5° = 95°

and a + 3d = 90° + 3×5° = 90° + 15°= 105°

Hence, the angles of the quadrilateral will ben75°, 85°, 95° and 105°.

**10. Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.**

**Solution**

Let the three parts of the number 207 are (a – d), a and (a + d), which are in A.P.

Now, by given condition,

⇒ Sum of these parts = 207

⇒ a – d + a + a + d = 207

⇒ 3a = 207

a = 69

Given that, product of the two smaller parts = 4623

⇒ a (a – d) = 4623

⇒ 69 (69 – d) = 4623

⇒ 69 – d = 67

⇒ d = 69 – 67 = 2

So, first part = a – d = 69 – 2 = 67,

Second part = a = 69

and third part = a + d = 69 + 2 = 71

Hence, required three parts are 67, 69, 71.

**11. The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles.**

**Solution**

Given that, the angles of a triangle are in A.P.

Let A, B and C are the angles of a ∆ABC.

2B = A+C** ...(i)**We know that, sum of all interior angles of a ∆ABC is 180°

A + B + C = 180°

⇒ 2B + B = 180°

**[from Eq. (i)]**

⇒ 3B = 180°

⇒ B = 60°

Let the greatest and least angles are A and C respectively.

A = 2C

**...(ii) [by condition]**

Now, put the values of B and A in Eq. (i), we get

2×60° = 2C + C

⇒ 120° = 3C

⇒ C = 40°

Put the value of C in Eq. (ii), we get

A = 2×40°

⇒ A = 80°

Hence, the required angles of triangle are 80°, 60° and 40°.

**12. The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number.**

**Solution**

Let the four consecutive numbers in A.P. be

a - 3d, a - d, a + d, a + 3d

So, a-3d + a-d + a+d + a+3d = 32

⇒ 4a = 32

⇒ a = 8

or, d = ± 2

So, when a = 8, d = 2, the numbers are 2, 6, 10, 14.