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 тАжequation (1)
a/r ├Ч a ├Ч ar = 3375 тАжequation (2)
From equation (2) weget,
a3┬а=3375
a = 15.
From equation (1) weget,
(a + ar + ar2)/r= 65
a + ar + ar2┬а=65r тАж equation (3)
Substituting a = 15 inequation (3) we get
15 + 15r + 15r2┬а=65r
15r2┬атАУ50r + 15 = 0тАж equation (4)
Dividing equation (4)by 5 we get
3r2┬атАУ10r + 3 = 0
3r2┬атАУ9r тАУ r + 3 = 0
3r(r тАУ 3) тАУ 1(r тАУ 3) =0
r = 3 or r = 1/3
Now, the equation willbe
15/3, 15, 15├Ч3 or
15/(1/3), 15, 15├Ч1/3
So the terms are 5,15, 45 or 45, 15, 5
тИ┤┬аThe 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 тАжequation (1)
a/r ├Ч a ├Ч ar = 1728 тАжequation (2)
From equation (2) weget,
a3┬а=1728
a = 12.
From equation (1) weget,
(a + ar + ar2)/r= 38
a + ar + ar2┬а=38r тАж equation (3)
Substituting a = 12 inequation (3) we get
12 + 12r + 12r2┬а=38r
12r2┬атАУ26r + 12 = 0тАж equation (4)
Dividing equation (4)by 2 we get
6r2┬атАУ13r + 6 = 0
6r2┬атАУ9r тАУ 4r + 6 = 0
3r(3r тАУ 3) тАУ 2(3r тАУ 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
тИ┤┬аThe threenumbers are 8, 12, 18
Question - 3 : - The sum of first three terms of a G.P. is 13/12, and their product is тАУ 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 тАжequation (1)
a/r ├Ч a ├Ч ar = -1 тАжequation (2)
From equation (2) weget,
a3┬а=-1
a = -1
From equation (1) weget,
(a + ar + ar2)/r= 13/12
12a + 12ar + 12ar2┬а=13r тАж equation (3)
Substituting a = тАУ 1in equation (3) we get
12( тАУ 1) + 12( тАУ 1)r +12( тАУ 1)r2┬а= 13r
12r2┬а+25r + 12 = 0
12r2┬а+16r + 9r + 12 = 0тАж equation (4)
4r (3r + 4) + 3(3r +4) = 0
r = -3/4┬аor r =-4/3
Now the equation willbe
-1/(-3/4), -1, -1├Ч-3/4or -1/(-4/3), -1, -1├Ч-4/3
4/3, -1, ┬╛ or ┬╛, -1,4/3
тИ┤┬аThe threenumbers are┬а4/3, -1, ┬╛ or ┬╛, -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 ┬╜ . Find them.
Answer - 4 : -
Let the three numbersbe a/r, a, ar
So, according to thequestion
a/r ├Ч a ├Ч ar = 125 тАжequation (1)
From equation (1) weget,
a3┬а=125
a = 5
a/r ├Ч a + a ├Ч ar + ar├Ч a/r = 87 ┬╜
a/r ├Ч a + a ├Ч ar + ar├Ч a/r = 195/2
a2/r + a2r+ a2┬а= 195/2
a2┬а(1/r+ r + 1) = 195/2
Substituting a = 5 inabove equation we get,
52┬а[(1+r2+r)/r]= 195/2
1+r2+r = (195r/2├Ч25)
2(1+r2+r) =39r/5
10 + 10r2┬а+10r = 39r
10r2┬атАУ29r + 10 = 0
10r2┬атАУ25r тАУ 4r + 10 = 0
5r(2r-5) тАУ 2(2r-5) = 0
r = 5/2, 2/5
So G.P is 10, 5, 5/2or 5/2, 5, 10
тИ┤┬аThe 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 тАжequation (1)
a/r ├Ч a ├Ч ar = 1 тАжequation (2)
From equation (2) weget,
a3┬а= 1
a = 1
From equation (1) weget,
(a + ar + ar2)/r= 39/10
10a + 10ar + 10ar2┬а=39r тАж equation (3)
Substituting a = 1 in3 we get
10(1) + 10(1)r +10(1)r2┬а= 39r
10r2┬атАУ29r + 10 = 0
10r2┬атАУ25r тАУ 4r + 10 = 0тАж equation (4)
5r(2r тАУ 5) тАУ 2(2r тАУ 5)= 0
r = 2/5 or 5/2
so now the equationwill be,
1/(2/5), 1, 1├Ч2/5 or1/(5/2), 1, 1├Ч5/2
5/2, 1, 2/5 or 2/5, 1,5/2
тИ┤┬аThe threenumbers are 2/5, 1, 5/2