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RD Chapter 19- Arithmetic Progressions Ex-19.6 Interview Questions Answers

Question 1 :

Find the A.M. between:
(i) 7 and 13 (ii) 12 and – 8 (iii) (x – y) and (x + y)

Answer 1 :

(i) Let A be theArithmetic mean

Then 7, A, 13 are inAP

Now, let us solve

A-7 = 13-A

2A = 13 + 7

A = 10

(ii) Let A be theArithmetic mean

Then 12, A, – 8 are inAP

Now, let us solve

A – 12 = – 8 – A

2A = 12 + 8

A = 2

(iii) Let A be theArithmetic mean

Then x – y, A, x + yare in AP

Now, let us solve

A – (x – y) = (x + y)– A

2A = x + y + x – y

A = x

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

Answer 2 :

Let A1, A2,A3, A4 be the 4 AM Between 4 and 19

Then, 4, A1,A2, A3, A4, 19 are in AP.

By using the formula,

d = (b-a) / (n+1)

= (19 – 4) / (4 + 1)

= 15/5

= 3

So,

A1 = a+ d = 4 + 3 = 7

A2 =A1 + d = 7 + 3 = 10

A3 =A2 + d = 10 + 3 = 13

A4 =A3 + d = 13 + 3 = 16

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

Answer 3 :

Let A1, A2,A3, A4, A5, A6, A7 bethe 7 AMs between 2 and 17

Then, 2, A1,A2, A3, A4, A5, A6, A7,17 are in AP

By using the formula,

an = a+ (n – 1)d

an =17, a = 2, n = 9

so,

17 = 2 + (9 – 1)d

17 = 2 + 9d – d

17 = 2 + 8d

8d = 17 – 2

8d = 15

d = 15/8

So,

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 2and 7 are 31/8, 46/8, 61/8, 76/8, 91/8, 106/8, 121/8

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

Answer 4 :

Let A1, A2,A3, A4, A5, A6 be the 7 AMbetween 15 and – 13

Then, 15, A1,A2, A3, A4, A5, A6, – 13are in AP

By using the formula,

an = a+ (n – 1)d

an =-13, a = 15, n = 8

so,

-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

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.

Answer 5 :

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

Given, an/a1 =3/1

We know total terms inAP are n + 2

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

By using the formula,

An = a+ (n – 1)d

An =17, a = 3

So, 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 inthe AP

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.

Answer 6 :

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

We know total terms inAP are n + 2

So 71 is the (n + 2)thterm

By using the formula,

An = a+ (n – 1)d

An =71, n = 6

A6 = a+ (6 – 1)d

a + 5d = 27 (5th term)

d = 4

so,

71 = (n + 2)th term

71 = a + (n + 2 – 1)d

71 = 7 + n(4)

n = 15

There are 15 terms inAP

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.

Answer 7 :

Let a and b be thefirst and last terms

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

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

Mean of A1 andAn = (A1 + An)/2

A1 =a+d

An = a– d

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

= (a+b)/2

AM between A2 andAn-1 = (a+2d+b-2d)/2

= (a+b)/2

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

Hence, AM = (a + b)/2

Question 8 :

If x, y, z are in A.P. and A1is the A.M. of x and y, and A2 isthe A.M. of y and z, then prove that the A.M. of A1 and A2 isy.

Answer 8 :

Given that,

A1 =AM of x and y

And A2 =AM of y and z

So, A1 =(x+y)/2

A2 =(y+x)/2

AM of A1 andA2 = (A1 + A2)/2

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

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

= [x+2y+z]/2

Since x, y, z are inAP, y = (x+z)/2

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

= (y + y)/2

= 2y/2

= y

Hence proved.

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

Answer 9 :

Let A1, A2,A3, A4, A5 be the 5 numbers between 8 and26

Then, 8, A1,A2, A3, A4, A5, 26 are in AP

By using the formula,

An = a+ (n – 1)d

An =26, a = 8, n = 7

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


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RD Chapter 19- Arithmetic Progressions Ex-19.6 Contributors

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