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Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.6
  • Last Updated : 15 Dec, 2020

Question 1: Find A.M. between:

(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.




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