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# Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.2

• Last Updated : 17 Dec, 2020

### Question 1. Find the three numbers in G.P whose sum is 65 and whose product is 3375.

Solution:

Let the three numbers in G.P as a/r, a, ar.

Given that product of three numbers is 3375.

Therefore,

a/r * a * ar = 3375

a3 = 3375

a3 = (15)3

Also, a/r + a + ar = 65

a(1/r + 1 + r) = 65

Substituting a = 15

15 (1/r + 1 + r) = 65

1/r + 1 + r = 65/15

1/r + 1 + r = 13/3

3r2 – 10r + 3 = 0

3r2 – 9r – r + 3 = 0

3r(r-3) – (r-3) = 0

(r-3)(3r-1) = 0

r = 3,1/3.

On taking r = 3, we get

a/r = 15 / 3 = 5 , a = 15 , ar = 15 * 3 = 45

On taking r = 1/3, we get

a/r = 15 * 3 = 45, a = 15, ar = 15 * 1/3 = 5

Therefore, the three terms are 5, 15, 45 or 45, 15, 5.

### Question 2. Find three numbers in G.P whose sum is 38 and their product is 1728.

Solution:

Let the three numbers in G.P as a/r, a, ar.

Given that product of three numbers is 1728.

Therefore,

a/r * a * ar = 1728

a3 = 1728

a3 = (12)3

Also, a/r + a + ar = 38

a (1 /r + 1 + r) = 38

Substituting a = 12

12 (1/r + 1 + r) = 38

1/r + 1 + r = 38/12

1/r + 1 + r = 19/6

6r2 – 13r + 6 = 0

6r2 – 9r – 4r + 6 = 0

3r(2r-3) – 2(2r-3) = 0

(2r-3)(3r-2) = 0

r = 3/2, 2/3.

On taking r = 3, we get

a/r = 12 * 2/3 = 8, a = 12, ar = 12 * 3/2 = 18

On taking r = 1/3, we get

a/r = 12 * 3/2 = 18 , a = 12, ar = 15 * 2/3 = 8

Therefore, the three terms are 8, 12, 18 or 18, 12, 8.

### Question 3. The sum of the first three terms of a G.P is 13/12 and their product is -1. Find the G.P

Solution:

Let the three numbers in G.P as a/r, a, ar.

Given that product of three numbers is -1.

Therefore, a/r * a * ar = -1

a3 = -1

a3 = (-1)3

a = -1

Also, a/r + a + ar = 13/12

a (1 /r + 1 + r) = 13/12

Substituting a = -1

-1 (1/r + 1 + r) = 13/12

1/r + 1 + r = -13/12

12r2 + 25r + 12 = 0

12r2 + 16r + 9r + 12 = 0

4r(3r+4) + 3(3r+4) = 0

(4r+3)(3r+4) = 0

r = -3/4, -4/3.

On taking r = -3/4, we get

a/r = -1 * -4/3 = 4/3, a = -1, ar = -1 * -3/4 = 3/4

On taking r = -4/3, we get

a/r = -1 * -3/4 = 3/4, a = -1, ar = -1 * -4/3 = 4/3

Therefore, the three terms are 4/3, -1, 3/4 or 3/4, -1, 4/3.

### Question 4. The product of three numbers in G.P is 125 and the sum of their product taken in pairs is 175/2. Find them.

Solution:

Let the three numbers in G.P as a/r, a, ar.

Given that product of three numbers is  125.

Therefore, a/r * a * ar = 125.

a3 = 125

a3 = (5)3

a = 5

Also,

a/r * a + a/r * ar + a*ar = 175/2

a2 (1 /r + 1 + r) = 175/2

Substituting a = 5

25 (1/r + 1 + r) = 175/2

1/r + 1 + r = 7/2

2r2 -5r + 2 = 0

2r2 -4r – r + 2= 0

2r(r – 2) – (r – 2) = 0

(2r-1)(r-2) = 0

r = 2, 1/2.

On taking r = 2, we get

a/r = 5/2, a = 2, ar = 5 * 2 = 10

On taking r = 1/2, we get

a/r = 5 * 2 = 10, a = 5, ar = 5 * 1/2 = 5/2

Therefore, the three terms are 5/2, 5, 10 or 10, 5, 5/2.

### Question 5. The sum of first three terms of a G.P is 39/10 and their product is 1. Find the common ratio and the terms.

Solution:

Assume the three numbers in G.P as a/r, a, ar.

Given that product of three numbers is 1.

Therefore, a/r * a * ar = 1

a3 = 1

a3 = (1)3

Also, a/r + a + ar = 39/10

a (1 /r + 1 + r) = 39/10

Substituting, a = 1,

1 (1/r + 1 + r) = 39/10

1/r + 1 + r = 39/10

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r-5) – 2(2r-5) = 0

(5r-2)(2r-5) = 0

r = 2/5, 5/2.

On taking r = 2/5, we get

a/r = 1 * 5/2 = 5/2, a = 1, ar = 1 * 2/5 = 2/5

On taking r = 5/2, we get

a/r = -1 * 2/5 = 2/5, a = 1, ar = 1 * 5/2 = 5/2

Therefore, the three terms are 5/2, 1, 2/5 or 2/5, 1, 5/2.

### Question 6. The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.

Solution:

Let the three numbers in G.P as a/r, a, ar.

First two terms are increased by 1 and third term is decreased by 1 , then it becomes A.P

a/r + 1, a + 1, ar – 1 is an A.P

Therefore,

ar – 1 – a – 1 = a – 1 – a/r – 1

ar – a – 2 = a – a/r

ar + a/r = 2a + 2 —(1)

Since, a/r + a + ar = 14

Replacing a/r + ar = 14 – a in equation (1)

14 – a = 2a + 2

12 = 3a

a = 4

Substituting a in equation (1)

r + 1/r = 5/2

2r2 – 5r + 2 = 0

2r2 – 4r -r + 2 = 0

2r(r-2) -(r-2) = 0

(r-2)(2r-1) = 0

r = 2, 1/2

On taking r = 2,

a/r = 4/2 = 2, a = 4, ar = 4*2 = 8

On taking r = 1/2

a/r = 4*2 = 8, a = 4 , ar = 4/2 = 2

Hence, the numbers are 2, 4, 8 or 8, 4, 2.

### Question 7. The product of three numbers in G.P. is 216. If 2,8,6 be added to them, the results are in A.P. Find the numbers.

Solution:

Let the three numbers be a/r, a, ar

Given that product of three number is 216.

Therefore,

a/r * a * ar = 216

a3 = (6)3

a = 6

Adding 2 on a/r, 8 on a and 6 on ar, it becomes an A.P

Therefore,

ar + 6 – a – 8 = a + 8 – a/r – 2,

Substituting a = 6

6r2 – 20r + 6 = 0

6r2 – 18r -2r + 6 = 0

6r(r-3) -2(r-3) = 0

r = 3, 1/3

On taking r = 3

a/r = 6/3 = 2, a = 6, ar = 6*3 = 18

On taking r = 1/3

a/r = 6*3 = 18, a = 6, ar = 6* 1/3 = 2

Hence, the numbers are 2, 6, 18 or 18, 6, 2.

### Question 8. Find three numbers in G.P. whose product is 729 and the sum of their products in pairs is 819.

Solution:

Let the three number be a/r, a, ar.

Given that product of three number is 729.

Therefore,

a/r * a * ar = 729

a3 = (9)3

a = 9

Also, sum of their products in pairs is 819.

a/r * a + a/r * ar + a * ar = 819

(a)2 * (1/r + 1 + r) = 819

Substituting a = 9

81 * (1/r + 1 + r) = 819

(1/r + 1 + r) = 91/9

9r2 – 82r + 9 = 0

9r2 – 81r – r + 9 = 0

9r(r-9) – (r-9) = 0

(r-9) (9r-1) = 0

r = 9, 1/9

On taking r = 9

a/r = 9/9 = 1, a = 9, ar = 9 * 9 = 81

On taking r = 1/9

a/r = 9 * 9 = 81, a = 9, ar = 9 * 1/9 = 1

Hence, the numbers are 1, 9, 81 or 81, 9, 1.

### Question 9. The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Solution:

Let the three numbers be a/r, a, ar

Given that a/r + a + ar = 21 —(1) and,

(a/r)2 + (a)2 + (ar)2 = 189 —(2)

We know that,

(A + B + C)2 = A2 + B2 +C2 + 2(AB + BC + CA)

Replacing A = a/r, B = a, C = ar,

21*21 = 189 + 2(a/r * a + a* ar + a*a)

126 = a2 [ 1/r + r + 1 ] —-(3)

From (1) we have 21 = a [ 1/r + r + 1] —(4)

Dividing (3) by (4), we get

a = 6

Substituting a in equation (4)

1/r + r + 1 = 7/2

2r2 – 5r + 2 = 0

2r2 – 4r – r + 2 = 0

2r(r-2) – (r-2) = 0

r = 2, 1/2

On taking r = 2

a/r = 3, a = 6, ar = 12

On taking r = 1/2

a/r = 12, a = 6, ar = 3

Hence, the numbers are 3, 6, 12 or 12, 6, 3

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