RD Chapter 20- Geometric Progressions Ex-20.1 |
RD Chapter 20- Geometric Progressions Ex-20.2 |
RD Chapter 20- Geometric Progressions Ex-20.3 |
RD Chapter 20- Geometric Progressions Ex-20.4 |
RD Chapter 20- Geometric Progressions Ex-20.5 |

**Answer
1** :

Let the six terms be a_{1},a_{2}, a_{3}, a_{4}, a_{5}, a_{6}.

A = 27, B = 1/81

Now, these 6terms are between A and B.

So the GP is: A, a_{1},a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, B.

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

We know that, T_{n} =ar^{n–1}

Here, T_{n} = 1/81,a = 27 and

1/81 = 27r^{8-1}

1/(81×27) = r^{7}

r = 1/3

a_{1} =Ar = 27×1/3 = 9

a_{2} =Ar^{2} = 27×1/9 = 3

a_{3} =Ar^{3} = 27×1/27 = 1

a_{4} =Ar^{4} = 27×1/81 = 1/3

a_{5} =Ar^{5} = 27×1/243 = 1/9

a_{6} =Ar^{6} = 27×1/729 = 1/27

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

**Answer
2** :

Let the five terms bea_{1}, a_{2}, a_{3}, a_{4}, a_{5}.

A = 27, B = 1/81

Now, these 5terms are between A and B.

So the GP is: A, a_{1},a_{2}, a_{3}, a_{4}, a_{5}, B.

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

We know that, T_{n} =ar^{n–1}

Here, T_{n} = 1/4,a = 16 and

1/4 = 16r^{7-1}

1/(4×16) = r^{6}

r = 1/2

a_{1} =Ar = 16×1/2 = 8

a_{2} =Ar^{2} = 16×1/4 = 4

a_{3} =Ar^{3} = 16×1/8 = 2

a_{4} =Ar^{4} = 16×1/16 = 1

a_{5} =Ar^{5} = 16×1/32 = 1/2

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

**Answer
3** :

Let the five terms bea_{1}, a_{2}, a_{3}, a_{4}, a_{5}.

A = 32/9, B = 81/2

Now, these 5terms are between A and B.

So the GP is: A, a_{1},a_{2}, a_{3}, a_{4}, a_{5}, B.

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

We know that, T_{n} =ar^{n–1}

Here, T_{n} = 81/2,a = 32/9 and

81/2 = 32/9r^{7-1}

(81×9)/(2×32) = r^{6}

r = 3/2

a_{1} =Ar = (32/9)×3/2 = 16/3

a_{2} =Ar^{2} = (32/9)×9/4 = 8

a_{3} =Ar^{3} = (32/9)×27/8 = 12

a_{4} =Ar^{4} = (32/9)×81/16 = 18

a_{5} =Ar^{5} = (32/9)×243/32 = 27

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

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a^{3}b and ab3

(iii) –8 and –2

**Answer
4** :

(i) 2 and 8

GM between a and b is√ab

Let a = 2 and b =8

GM = √2×8

= √16

= 4

(ii) a^{3}band ab^{3}

GM between a and b is√ab

Let a = a^{3}band b = ab^{3}

GM = √(a^{3}b× ab^{3})

= √a^{4}b^{4}

= a^{2}b^{2}

(iii) –8 and –2

GM between a and b is√ab

Let a = –2 and b = –8

GM = √(–2×–8)

= √–16

= -4

If a is the G.M. of 2 and ¼ find a.

**Answer
5** :

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

**Answer
6** :

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 – b^{2} =400

b^{2} –50b + 400 = 0

b^{2} –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.

**Answer
7** :

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 = G^{2}…(2)

The quadratic equationis given by,

x^{2 }– x (Sumof roots) + (Product of roots) = 0

x^{2} – x (2A)+ (G^{2}) = 0

x^{2} –2Ax + G^{2} = 0 [Using (1) and (2)]

∴ The requiredquadratic equation is x^{2} – 2Ax + G^{2} =0.

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