# Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.3

• Last Updated : 03 Jan, 2021

### Question 1. The sum of first three terms of an AP is 21 and the product of first and the third term exceed the second term by 6, find three terms.

Solution:

We are given that the sum of first three terms are 21. Let’s suppose these three terms are a-d, a, a+d.

So,

a-d + a + a+d = 21

3a = 21

a= 7

And we are given

(a-d)*(a+d) – a = 6

a2 – d2 -a = 6

Putting value of a in this equation we can get value of d.

(7)2 – (d)2 -7 = 6

49 – d2 – 7 = 6

d2 = 36

d= ±6

For d = 6 three terms of AP are 1, 7, 13.

For d = -6 three terms are 13, 7, 1.

### Question 2. Three numbers are in AP. If sum of these numbers be 27 and the product 648, find the numbers.

Solution:

We are given that the sum of first three terms are 27. Let’s suppose these three terms are a-d, a, a+d.

So

a-d + a + a+d = 27

3a = 27

a= 9

And we are given

(a-d)*a*(a+d)  =648

Putting value of a in this equation we can get value of d.

93– 9d2 =648

729- 9d2 =648

9d2 = 81

d2 = 9

d= ±3

For d = 3 three terms of AP are 6, 9, 12.

For d = -3 three terms are 12, 9, 6.

### Question 3. Find the four numbers in AP, whose sum is 50 and in which the greatest number is 4 times the least.

Solution:

We are given that the sum of first four terms are 50.  Let’s suppose these four terms are a-3d, a-d, a+d, a+3d.

So

a-3d + a-d + a+d + a+3d = 50

4a = 50

a = 25/2

And we are given greatest number is 4 times the least,

4(a-3d) = a +3d

4a – 12d = a + 3d

3a = 15d

a = 5d

Putting value of a, we get

d = a/5 = 5/2

So the four terms of AP are

a – 3d = 25/2- 3*5/2 = 5

a – d  = 25/2- 5/2 = 10

a + d  = 25/2 + 5/2 = 15

a + 3d = 25/2 + 3*5/2 = 20

### Question 4. The sum of three numbers in AP is 12, and sum of there cubes is 288. Find the numbers.

Solution:

We are given that the sum of first three terms are 12. Let’s suppose these three terms are a-d, a, a+d.

So

a-d + a + a+d = 12

3a = 12

a= 4

And we are given sum of there cubes is 288.

(a-d)3 + a3 + (a+d)3 = 288

Putting value of a in this equation,

43-d3-3*4*d(4-d) + 43 + 43 + d3 + 3*4*d(4+d) = 288

64- d3-12d(4-d) + 64+64+ d3+ 12d(4+d) = 288

192+24d2 =288

24d2=96

d2= 4

d= ±2

For d = 2 three terms of AP are 2, 4, 6.

For d = -2 three terms are 6, 4, 2.

### Question 5. If the sum of three numbers in AP is 24 and their product is 440. Find the numbers.

Solution:

We are given that the sum of first three terms are 24. Let’s suppose these three terms are a-d, a, a+d.

So

a-d + a + a +d = 24

3a = 24

a = 8

And we are given

(a-d)*a*(a+d) =440

Putting value of a in this equation we can get value of d.

83– 8d2 = 440

512- 8d2 = 440

8d2 = 72

d2 = 9

d = ±3

For d = 3 three terms of AP are 5, 8, 11.

For d = -3 three terms are 11, 8, 5.

### Question 6. The angles of a quadrilateral are in AP whose common difference is 10. Find the angles.

Solution:

We are given that the angles of a quadrilateral are in AP and we know that there are four angles in quadrilateral and sum of all angles in 360.

So, let’s suppose these four angles are a-3d, a-d, a+d, a+3d.

We know sum of these angles is 360.

a-3d + a-d + a+d + a+3d = 360

4a = 360

a =90

We are given that common difference is 10.

(a-d) – (a-3d) = 10

2d = 10

d = 5

So the four angles are

a-3d = 90-15= 75

a-d = 90 – 5 = 85

a+d = 90 + 5 = 95

a+3d = 90+15 = 105

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