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

**Question 1: Find A.M. between:**

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

**Solution: **

(i) 7 and 13Let A be the Arithmetic Mean of 7 and 13.

Then,

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

A – 7 = 13 – A2A = 13 + 7

A = 20/2 = 10

∴ A.M. = 10

(ii) 12 and -8Let 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 A

_{1}, A_{2}, A_{3}, A_{4}be the 4 A.M.s Between 4 and 19.Then, 4, A

_{1}, A_{2}, A_{3}, A_{4}, 19 are in A.P. of 6 terms.We know,

A

_{n }= a + (n – 1).da = 4

Then,

a

_{6}= 19 = 4 + (6 – 1).d∴ d = 3

Now,

A

_{1}= a + d = 4 + 3 = 7A

_{2}= A_{1}+ d = 7 + 3 = 10A

_{3}= A_{2}+ d = 10 + 3 = 13A

_{4}= A_{3}+ 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 A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}be the 7 A.M.s between 2 and 17.Then, 2, A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, 17 are in A.P. of 9 terms.We know,

A

_{n}= a + (n – 1).da = 2

Then,

a

_{9}= 17 = 2 + (9 – 1).d17 = 2 + 9d – d

17 = 2 + 8d

8d = 17 – 2

8d = 15

∴ d = 15/8

Now,

A

_{1}= a + d = 2 + 15/8 = 31/8A

_{2}= A_{1}+ d = 31/8 + 15/8 = 46/8A

_{3}= A_{2}+ d = 46/8 + 15/8 = 61/8A

_{4}= A_{3}+ d = 61/8 + 15/8 = 76/8A

_{5}= A_{4}+ d = 76/8 + 15/8 = 91/8A

_{6}= A_{5}+ d = 91/8 + 15/8 = 106/8A

_{7}= A_{6}+ 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 A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}be the 6 A.M.s between 15 and –13.Then, 15, A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, –13 are in A.P. series.We know,

An = a + (n – 1)d

a = 15

Then,

a

_{8}= -13 = 15 + (8 – 1)d-13 = 15 + 7d

7d = -13 – 15

7d = -28

∴ d = -4

So,

A

_{1}= a + d = 15 – 4 = 11A

_{2}= A_{1}+ d = 11 – 4 = 7A

_{3}= A_{2}+ d = 7 – 4 = 3A

_{4}= A_{3}+ d = 3 – 4 = -1A

_{5}= A_{4}+ d = -1 – 4 = -5A

_{6}= A_{5}+ 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, A

_{1}, A_{2}, A_{3}, …….., A_{n}, 17.Given,

A

_{n}/A_{1}= 3/1We know total terms in AP are n + 2.

So, 17 is the (n + 2)

^{th}term.We know,

A

_{n}= a + (n – 1)da = 3

Then,

A

_{n}= 17, a = 3A

_{n}= 17 = 3 + (n + 2 – 1)d17 = 3 + (n + 1)d

17 – 3 = (n + 1)d

14 = (n + 1)d

∴ d = 14/(n + 1).

Now,

A

_{n}= 3 + 14/(n + 1) = (17n + 3)/(n + 1)A

_{1}= 3 + d = (3n + 17)/(n + 1)Since,

A

_{n}/A_{1}= 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, A

_{1}, A_{2}, A_{3}, …….., A_{n}, 71We know total terms in the A.P. series is n + 2.

So, 71 is the (n + 2)

^{th}termWe know,

A

_{n}= a + (n – 1)da = 7

Then,

5

^{th}A.M. = A6 = a + (6 – 1)da + 5d = 27

∴ d = 4

So,

71 = (n + 2)

^{th}term71 = 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, A

_{1}, A_{2}, A_{3}, …….., A_{n}, b.We know, Mean of a and b = (a + b)/2

Mean of A

_{1}and A_{n}= (A_{1}+ A_{n})/2∴ A

_{1}= a + d∴ A

_{n}= a – dSo, AM = (a + d + b – d) / 2 = (a + b) / 2

A.M. between A

_{2}and A_{n-1}= (a + 2d + b – 2d) / 2 = (a + b) / 2Similarly, (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,

A

_{1}= A.M. of x and yA

_{2}= A.M. of y and zSo,

A

_{1}= (x + y)/2A

_{2}= (y + z)/2AM of A

_{1}and A_{2}= (A_{1}+ A_{2})/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 A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}be the 5 numbers between 8 and 26Then, 8, A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26 are in the A.P. seriesWe know,

A

_{n}= a + (n – 1)da = 8

Then,

a

_{7}= 26 = 8 + (7 – 1)d26 = 8 + 6d

6d = 26 – 8

6d = 18

∴ d = 18/6 = 3

So,

A

_{1}= a + d = 8 + 3 = 11A

_{2}= A_{1}+ d = 11 + 3 = 14A

_{3}= A_{2}+ d = 14 + 3 = 17A

_{4}= A_{3}+ d = 17 + 3 = 20A

_{5}= A_{4}+ d = 20 + 3 = 23∴ So, the A.P. series is 8, 11, 14, 17, 20, 23 and 26.

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