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

**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.Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

Demo Class for First Step to Coding Course,specificallydesigned for students of class 8 to 12.The students will get to learn more about the world of programming in these

free classeswhich will definitely help them in making a wise career choice in the future.Given that product of three numbers is 3375.

Therefore,

a/r * a * ar = 3375

a

^{3}= 3375a

^{3}= (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

3r

^{2}– 10r + 3 = 03r

^{2}– 9r – r + 3 = 03r(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 = 45On taking

r = 1/3, we get

a/r = 15 * 3 = 45, a = 15, ar = 15 * 1/3 = 5Therefore, the three terms are

5, 15, 45or45, 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

a

^{3}= 1728a

^{3}= (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

6r

^{2}– 13r + 6 = 06r

^{2}– 9r – 4r + 6 = 03r(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 = 18On taking

r = 1/3, we get

a/r = 12 * 3/2 = 18 , a = 12, ar = 15 * 2/3 = 8Therefore, the three terms are

8, 12, 18or18, 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

a

^{3}= -1a

^{3}= (-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

12r

^{2}+ 25r + 12 = 012r

^{2}+ 16r + 9r + 12 = 04r(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 = 3/4On taking

r = -4/3, we get

a/r = -1 * -3/4 = 3/4, a = -1, ar = -1 * -4/3 = 4/3Therefore, the three terms are

4/3, -1, 3/4or3/4, -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.

a

^{3}= 125a

^{3}= (5)^{3}

a = 5Also,

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

a

^{2}(1 /r + 1 + r) = 175/2Substituting a = 5

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

1/r + 1 + r = 7/2

2r

^{2}-5r + 2 = 02r

^{2}-4r – r + 2= 02r(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 = 10On taking

r = 1/2, we get

a/r = 5 * 2 = 10, a = 5, ar = 5 * 1/2 = 5/2Therefore, the three terms are

5/2, 5, 10or10, 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

a

^{3}= 1a

^{3}= (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

10r

^{2}– 29r + 10 = 010r

^{2}– 25r – 4r + 10 = 05r(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 = 2/5On taking

r = 5/2, we get

a/r = -1 * 2/5 = 2/5, a = 1, ar = 1 * 5/2 = 5/2Therefore, the three terms are

5/2, 1, 2/5or2/5, 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

2r

^{2}– 5r + 2 = 02r

^{2}– 4r -r + 2 = 02r(r-2) -(r-2) = 0

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

r = 2, 1/2On taking

r = 2,

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

r = 1/2

a/r = 4*2 = 8, a = 4 , ar = 4/2 = 2Hence, the numbers are

2, 4, 8or8, 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

a

^{3}= (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

6r

^{2}– 20r + 6 = 06r

^{2}– 18r -2r + 6 = 06r(r-3) -2(r-3) = 0

r = 3, 1/3On taking

r = 3

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

r = 1/3

a/r = 6*3 = 18, a = 6, ar = 6* 1/3 = 2Hence, the numbers are

2, 6, 18or18, 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

a

^{3}= (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) = 819Substituting a = 9

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

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

9r

^{2}– 82r + 9 = 09r

^{2}– 81r – r + 9 = 09r(r-9) – (r-9) = 0

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

r = 9, 1/9On taking

r = 9

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

r = 1/9

a/r = 9 * 9 = 81, a = 9, ar = 9 * 1/9 = 1Hence, the numbers are

1, 9, 81or81, 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, arGiven that a/r + a + ar = 21 —(1) and,

(a/r)

^{2}+ (a)^{2}+ (ar)^{2}= 189 —(2)We know that,

(A + B + C)

^{2}= A^{2}+ B^{2}+C^{2}+ 2(AB + BC + CA)Replacing A = a/r, B = a, C = ar,

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

126 = a

^{2}[ 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

2r

^{2}– 5r + 2 = 02r

^{2}– 4r – r + 2 = 02r(r-2) – (r-2) = 0

r = 2, 1/2On taking

r = 2

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

r = 1/2

a/r = 12, a = 6, ar = 3Hence, the numbers are

3, 6, 12or12, 6, 3