RD Chapter 20 Geometric Progressions Ex 20.2 Solutions
Question - 1 : - Find three numbers in G.P. whose sum is 65 andwhose product is 3375.
Answer - 1 : -
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.
Answer - 2 : -
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.
Answer - 3 : -
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.
Answer - 4 : -
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.
Answer - 5 : -
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