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# RD Chapter 20- Geometric Progressions Ex-20.6 Interview Questions Answers

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Question 1 : Insert 6 geometric means between 27 and 1/81.

Let the six terms be a1,a2, a3, a4, a5, a6.

A = 27, B = 1/81

Now, these 6terms are between A and B.

So the GP is: A, a1,a2, a3, a4, a5, a6, B.

So we now have 8 termsin GP with the first term being 27 and eighth being 1/81.

We know that, Tn =arn–1

Here, Tn = 1/81,a = 27 and

1/81 = 27r8-1

1/(81×27) = r7

r = 1/3

a1 =Ar = 27×1/3 = 9

a2 =Ar2 = 27×1/9 = 3

a3 =Ar3 = 27×1/27 = 1

a4 =Ar4 = 27×1/81 = 1/3

a5 =Ar5 = 27×1/243 = 1/9

a6 =Ar6 = 27×1/729 = 1/27

The six GMbetween 27 and 1/81 are 9, 3, 1, 1/3, 1/9, 1/27

Question 2 : Insert 5 geometric means between 16 and 1/4.

Let the five terms bea1, a2, a3, a4, a5.

A = 27, B = 1/81

Now, these 5terms are between A and B.

So the GP is: A, a1,a2, a3, a4, a5, B.

So we now have 7 termsin GP with the first term being 16 and seventh being 1/4.

We know that, Tn =arn–1

Here, Tn = 1/4,a = 16 and

1/4 = 16r7-1

1/(4×16) = r6

r = 1/2

a1 =Ar = 16×1/2 = 8

a2 =Ar2 = 16×1/4 = 4

a3 =Ar3 = 16×1/8 = 2

a4 =Ar4 = 16×1/16 = 1

a5 =Ar5 = 16×1/32 = 1/2

The five GMbetween 16 and 1/4 are 8, 4, 2, 1, ½

Question 3 : Insert 5 geometric means between 32/9 and 81/2.

Let the five terms bea1, a2, a3, a4, a5.

A = 32/9, B = 81/2

Now, these 5terms are between A and B.

So the GP is: A, a1,a2, a3, a4, a5, B.

So we now have 7 termsin GP with the first term being 32/9 and seventh being 81/2.

We know that, Tn =arn–1

Here, Tn = 81/2,a = 32/9 and

81/2 = 32/9r7-1

(81×9)/(2×32) = r6

r = 3/2

a1 =Ar = (32/9)×3/2 = 16/3

a2 =Ar2 = (32/9)×9/4 = 8

a3 =Ar3 = (32/9)×27/8 = 12

a4 =Ar4 = (32/9)×81/16 = 18

a5 =Ar5 = (32/9)×243/32 = 27

The five GMbetween 32/9 and 81/2 are 16/3, 8, 12, 18, 27

Question 4 :
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2

(i) 2 and 8

GM between a and b is√ab

Let a = 2 and b =8

GM = √2×8

= √16

= 4

(ii) a3band ab3

GM between a and b is√ab

Let a = a3band b = ab3

GM = √(a3b× ab3)

= √a4b4

= a2b2

(iii) –8 and –2

GM between a and b is√ab

Let a = –2 and b = –8

GM = √(–2×–8)

= √–16

= -4

Question 5 :

If a is the G.M. of 2 and ¼ find a.

We know that GMbetween a and b is √ab

Let a = 2 and b = 1/4

GM = √(2×1/4)

= √(1/2)

= 1/√2

value of a is 1/√2

Question 6 : Find the two numbers whose A.M. is 25 and GM is 20.

Given: A.M = 25, G.M =20.

G.M = √ab

A.M = (a+b)/2

So,

√ab = 20 ……. (1)

(a+b)/2 = 25……. (2)

a + b = 50

a = 50 – b

Putting the value of‘a’ in equation (1), we get,

√[(50-b)b] = 20

50b – b2 =400

b2 –50b + 400 = 0

b2 –40b – 10b + 400 = 0

b(b – 40) – 10(b – 40)= 0

b = 40 or b = 10

If b = 40 then a = 10

If b = 10 then a = 40

The numbers are 10and 40.

Question 7 : Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Let the root of thequadratic equation be a and b.

So, according to thegiven condition,

A.M = (a+b)/2 = A

a + b = 2A ….. (1)

GM = √ab = G

ab = G2…(2)

x– x (Sumof roots) + (Product of roots) = 0

x2 – x (2A)+ (G2) = 0

x2 –2Ax + G2 = 0 [Using (1) and (2)]

The requiredquadratic equation is x2 – 2Ax + G2 =0.

krishan