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# RD Chapter 20- Geometric Progressions Ex-20.2 Interview Questions Answers

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Question 1 : Find three numbers in G.P. whose sum is 65 andwhose product is 3375.

Let the three numbersbe a/r, a, ar

So, according to thequestion

a/r + a + ar = 65 …equation (1)

a/r × a × ar = 3375 …equation (2)

From equation (2) weget,

a3 =3375

a = 15.

From equation (1) weget,

(a + ar + ar2)/r= 65

a + ar + ar2 =65r … equation (3)

Substituting a = 15 inequation (3) we get

15 + 15r + 15r2 =65r

15r2 –50r + 15 = 0… equation (4)

Dividing equation (4)by 5 we get

3r2 –10r + 3 = 0

3r2 –9r – r + 3 = 0

3r(r – 3) – 1(r – 3) =0

r = 3 or r = 1/3

Now, the equation willbe

15/3, 15, 15×3 or

15/(1/3), 15, 15×1/3

So the terms are 5,15, 45 or 45, 15, 5

The threenumbers are 5, 15, 45.

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

Let the three numbersbe a/r, a, ar

So, according to thequestion

a/r + a + ar = 38 …equation (1)

a/r × a × ar = 1728 …equation (2)

From equation (2) weget,

a3 =1728

a = 12.

From equation (1) weget,

(a + ar + ar2)/r= 38

a + ar + ar2 =38r … equation (3)

Substituting a = 12 inequation (3) we get

12 + 12r + 12r2 =38r

12r2 –26r + 12 = 0… equation (4)

Dividing equation (4)by 2 we get

6r2 –13r + 6 = 0

6r2 –9r – 4r + 6 = 0

3r(3r – 3) – 2(3r – 3)= 0

r = 3/2 or r = 2/3

Now the equation willbe

12/(3/2) = 8 or

12/(2/3) = 18

So the terms are 8,12, 18

The threenumbers are 8, 12, 18

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

Let the three numbersbe a/r, a, ar

So, according to thequestion

a/r + a + ar = 13/12 …equation (1)

a/r × a × ar = -1 …equation (2)

From equation (2) weget,

a3 =-1

a = -1

From equation (1) weget,

(a + ar + ar2)/r= 13/12

12a + 12ar + 12ar2 =13r … equation (3)

Substituting a = – 1in equation (3) we get

12( – 1) + 12( – 1)r +12( – 1)r2 = 13r

12r2 +25r + 12 = 0

12r2 +16r + 9r + 12 = 0… equation (4)

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

r = -3/4 or r =-4/3

Now the equation willbe

-1/(-3/4), -1, -1×-3/4or -1/(-4/3), -1, -1×-4/3

4/3, -1, ¾ or ¾, -1,4/3

The threenumbers are 4/3, -1, ¾ or ¾, -1, 4/3

Question 4 : The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is 87 ½ . Find them.

Let the three numbersbe a/r, a, ar

So, according to thequestion

a/r × a × ar = 125 …equation (1)

From equation (1) weget,

a3 =125

a = 5

a/r × a + a × ar + ar× a/r = 87 ½

a/r × a + a × ar + ar× a/r = 195/2

a2/r + a2r+ a2 = 195/2

a2 (1/r+ r + 1) = 195/2

Substituting a = 5 inabove equation we get,

52 [(1+r2+r)/r]= 195/2

1+r2+r = (195r/2×25)

2(1+r2+r) =39r/5

10 + 10r2 +10r = 39r

10r2 –29r + 10 = 0

10r2 –25r – 4r + 10 = 0

5r(2r-5) – 2(2r-5) = 0

r = 5/2, 2/5

So G.P is 10, 5, 5/2or 5/2, 5, 10

The threenumbers are 10, 5, 5/2 or 5/2, 5, 10

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

Let the three numbersbe a/r, a, ar

So, according to thequestion

a/r + a + ar = 39/10 …equation (1)

a/r × a × ar = 1 …equation (2)

From equation (2) weget,

a3 = 1

a = 1

From equation (1) weget,

(a + ar + ar2)/r= 39/10

10a + 10ar + 10ar2 =39r … equation (3)

Substituting a = 1 in3 we get

10(1) + 10(1)r +10(1)r2 = 39r

10r2 –29r + 10 = 0

10r2 –25r – 4r + 10 = 0… equation (4)

5r(2r – 5) – 2(2r – 5)= 0

r = 2/5 or 5/2

so now the equationwill be,

1/(2/5), 1, 1×2/5 or1/(5/2), 1, 1×5/2

5/2, 1, 2/5 or 2/5, 1,5/2

The threenumbers are 2/5, 1, 5/2

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