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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 

                                 a3 – ad = 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

                 a3-d3-3ad(a-d) +a3+ a3 +d3 + 3ad(a+d) = 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

                                a3 – ad2  = 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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