Sometimes it is required to select a finite number of terms in G.P. If the product of the numbers is not given, then the numbers are taken as a, ar… In this exercise, we mainly study problems based on a selection of terms in G.P. Examples are provided, which illustrate the application of a selection of terms. Experts at BYJU’S have solved the exercise problems using shortcut techniques to help students increase the speed of solving the problems. Students can also refer to RD Sharma Class 11 Maths Solutions, and by practising regularly, they can achieve their goals of scoring high marks. These solutions are available in PDF format, students can download them and can learn offline.
RD Sharma Solutions for Class 11 Maths Exercise 20.2 Chapter 20 – Geometric Progression
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1. Find three numbers in G.P. whose sum is 65 and whose product is 3375.
Solution:
Let the three numbers be a/r, a, ar
So, according to the question
a/r + a + ar = 65 … equation (1)
a/r × a × ar = 3375 … equation (2)
From equation (2), we get,
a3 = 3375
a = 15.
From equation (1), we get,
(a + ar + ar2)/r = 65
a + ar + ar2 = 65r … equation (3)
Substituting a = 15 in equation (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 will be
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 three numbers are 5, 15, 45.
2. Find three number in G.P. whose sum is 38 and their product is 1728.
Solution:
Let the three numbers be a/r, a, ar
So, according to the question
a/r + a + ar = 38 … equation (1)
a/r × a × ar = 1728 … equation (2)
From equation (2) we get,
a3 = 1728
a = 12.
From equation (1), we get,
(a + ar + ar2)/r = 38
a + ar + ar2 = 38r … equation (3)
Substituting a = 12 in equation (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 will be
12/(3/2) = 8 or
12/(2/3) = 18
So the terms are 8, 12, 18
∴ The three numbers are 8, 12, 18
3. The sum of first three terms of a G.P. is 13/12, and their product is – 1. Find the G.P.
Solution:
Let the three numbers be a/r, a, ar
So, according to the question
a/r + a + ar = 13/12 … equation (1)
a/r × a × ar = -1 … equation (2)
From equation (2), we get,
a3 = -1
a = -1
From equation (1), we get,
(a + ar + ar2)/r = 13/12
12a + 12ar + 12ar2 = 13r … equation (3)
Substituting a = – 1 in 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 will be
-1/(-3/4), -1, -1×-3/4 or -1/(-4/3), -1, -1×-4/3
4/3, -1, ¾ or ¾, -1, 4/3
∴ The three numbers are 4/3, -1, ¾ or ¾, -1, 4/3
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.
Solution:
Let the three numbers be a/r, a, ar
So, according to the question
a/r × a × ar = 125 … equation (1)
From equation (1), we get,
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 in the above 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, 5/2 or 5/2, 5, 10
∴ The three numbers are 10, 5, 5/2 or 5/2, 5, 10
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.
Solution:
Let the three numbers be a/r, a, ar
So, according to the question
a/r + a + ar = 39/10 … equation (1)
a/r × a × ar = 1 … equation (2)
From equation (2) we get,
a3 = 1
a = 1
From equation (1), we get,
(a + ar + ar2)/r = 39/10
10a + 10ar + 10ar2 = 39r … equation (3)
Substituting a = 1 in 3, 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 equation will be,
1/(2/5), 1, 1×2/5 or 1/(5/2), 1, 1×5/2
5/2, 1, 2/5 or 2/5, 1, 5/2
∴ The three numbers are 2/5, 1, 5/2
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