Related Articles

# Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.6

• Last Updated : 15 Dec, 2020

### (i) 7 and 13 (ii) 12 and -8 (iii) (x – y) and (x + y)

Solution:

(i) 7 and 13

Let A be the Arithmetic Mean of 7 and 13.

Then,
7, A and 13 are in A.P.

Now,
A – 7 = 13 – A

2A = 13 + 7

A = 20/2 = 10

∴ A.M. = 10

(ii) 12 and -8

Let A be the Arithmetic Mean of 12 and -8.

Then, 12, A and -8 are in A.P.

Now,

A – 12 = – 8 – A

2A = 12 + 8

A = 2

∴ A.M. = 2

(iii) (x – y) and (x + y)

Let A be the Arithmetic Mean of (x – y) and (x + y).

Then, (x – y), A and (x + y) are in A.P.

Now,

A – (x – y) = (x + y) – A

2A = x + y + x – y

A = x

∴ A.M. = x

### Question 2: Insert 4 A.M.s between 4 and 19

Solution:

Let A1, A2, A3, A4 be the 4 A.M.s Between 4 and 19.

Then, 4, A1, A2, A3, A4, 19 are in A.P. of 6 terms.

We know,

An = a + (n – 1).d

a = 4

Then,

a6 = 19 = 4 + (6 – 1).d

∴ d = 3

Now,

A1 = a + d = 4 + 3 = 7

A2 = A1 + d = 7 + 3 = 10

A3 = A2 + d = 10 + 3 = 13

A4 = A3 + d = 13 + 3 = 16

∴ The 4 A.M.s between 4 and 16 are 7, 10, 13 and 16.

### Question 3: Insert 7 A.M.s between 2 and 17

Solution:

Let A1, A2, A3, A4, A5, A6, A7 be the 7 A.M.s between 2 and 17.

Then, 2, A1, A2, A3, A4, A5, A6, A7, 17 are in A.P. of 9 terms.

We know,

An = a + (n – 1).d

a = 2

Then,

a9 = 17 = 2 + (9 – 1).d

17 = 2 + 9d – d

17 = 2 + 8d

8d = 17 – 2

8d = 15

∴ d = 15/8

Now,

A1 = a + d = 2 + 15/8 = 31/8

A2 = A1 + d = 31/8 + 15/8 = 46/8

A3 = A2 + d = 46/8 + 15/8 = 61/8

A4 = A3 + d = 61/8 + 15/8 = 76/8

A5 = A4 + d = 76/8 + 15/8 = 91/8

A6 = A5 + d = 91/8 + 15/8 = 106/8

A7 = A6 + d = 106/8 + 15/8 = 121/8

∴ The 7 AMs between 2 and 7 are 31/8, 46/8, 61/8, 76/8, 91/8, 106/8 and 121/8.

### Question 4: Insert six A.M.s between 15 and – 13

Solution:

Let A1, A2, A3, A4, A5, A6 be the 6 A.M.s between 15 and –13.

Then, 15, A1, A2, A3, A4, A5, A6, –13 are in A.P. series.

We know,

An = a + (n – 1)d

a = 15

Then,

a8 = -13 = 15 + (8 – 1)d

-13 = 15 + 7d

7d = -13 – 15

7d = -28

∴ d = -4

So,

A1 = a + d = 15 – 4 = 11

A2 = A1 + d = 11 – 4 = 7

A3 = A2 + d = 7 – 4 = 3

A4 = A3 + d = 3 – 4 = -1

A5 = A4 + d = -1 – 4 = -5

A6 = A5 + d = -5 – 4 = -9

∴ The 6 A.M.s between 15 and -13 are 11, 7, 3, -1, -5 and -9.

### Question 5: There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n.

Solution:

Let the A.P. series be 3, A1, A2, A3, …….., An, 17.

Given,

An/A1 = 3/1

We know total terms in AP are n + 2.

So, 17 is the (n + 2)th term.

We know,

An = a + (n – 1)d

a = 3

Then,

An = 17, a = 3

An = 17 = 3 + (n + 2 – 1)d

17 = 3 + (n + 1)d

17 – 3 = (n + 1)d

14 = (n + 1)d

∴ d = 14/(n + 1).

Now,

An = 3 + 14/(n + 1) = (17n + 3)/(n + 1)

A1 = 3 + d = (3n + 17)/(n + 1)

Since,

An/A1 = 3/1

(17n + 3)/(3n + 17) = 3/1

17n + 3 = 3(3n + 17)

17n + 3 = 9n + 51

17n – 9n = 51 – 3

8n = 48

n = 48/8

= 6

∴ There are 6 terms in the A.P. series.

### Question 6: Insert A.M.s between 7 and 71 in such a way that the 5th A.M. is 27. Find the number of A.M.s.

Solution:

Let the series be 7, A1, A2, A3, …….., An, 71

We know total terms in the A.P. series is n + 2.

So, 71 is the (n + 2)th term

We know,

An = a + (n – 1)d

a = 7

Then,

5th A.M. = A6 = a + (6 – 1)d

a + 5d = 27

∴ d = 4

So,

71 = (n + 2)th term

71 = a + (n + 2 – 1).d

71 = 7 + n.(4)

n = 15

∴ There are 15 terms in the A.P. series.

### Question 7: If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.

Solution:

Let a and b be the first and last terms.

The series be a, A1, A2, A3, …….., An, b.

We know, Mean of a and b = (a + b)/2

Mean of A1 and An = (A1 + An)/2

∴ A1 = a + d

∴ An = a – d

So, AM = (a + d + b – d) / 2 = (a + b) / 2

A.M. between A2 and An-1 = (a + 2d + b – 2d) / 2 = (a + b) / 2

Similarly, (a + b) / 2 is constant for all such numbers.

∴ Hence, A.M. = (a + b) / 2

### Question 8: If x, y, z are in A.P. and A1is the A.M. of x and y, and A2 is the A.M. of y and z, then prove that the A.M. of A1 and A2 is y.

Solution:

Given that,

A1 = A.M. of x and y

A2 = A.M. of y and z

So,

A1 = (x + y)/2

A2 = (y + z)/2

AM of A1 and A2 = (A1 + A2)/2

= [(x + y)/2 + (y + z)/2]/2

= (x + y + y + z)/4

= (x + 2y + z)/4  ……(i)

Since x, y, z are in AP,

y = (x + z)/2  ……(ii)

From (i) and (ii),

A.M. = [{x + z/2} + {(x + 2y + z)/4}]/2

= (y + y)/2

= 2y/2 = y

A.M. = y [Hence proved]

### Question 9: Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Solution:

Let A1, A2, A3, A4, A5 be the 5 numbers between 8 and 26

Then, 8, A1, A2, A3, A4, A5, 26 are in the A.P. series

We know,

An = a + (n – 1)d

a = 8

Then,

a7 = 26 = 8 + (7 – 1)d

26 = 8 + 6d

6d = 26 – 8

6d = 18

∴ d = 18/6 = 3

So,

A1 = a + d = 8 + 3 = 11

A2 = A1 + d = 11 + 3 = 14

A3 = A2 + d = 14 + 3 = 17

A4 = A3 + d = 17 + 3 = 20

A5 = A4 + d = 20 + 3 = 23

∴ So, the A.P. series is 8, 11, 14, 17, 20, 23 and 26.

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

My Personal Notes arrow_drop_up