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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

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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