RD Chapter 20 Geometric Progressions Ex 20.6 Solutions
Question - 1 : - Insert 6 geometric means between 27 and 1/81.
Answer - 1 : -
Let the six terms be a1,a2, a3, a4, a5, a6.
A = 27, B = 1/81
Now,┬аthese 6terms are between A and B.
So the GP is: A, a1,a2, a3, a4, a5, a6, B.
So we now have 8 termsin GP with the first term being 27 and eighth being 1/81.
We know that,┬аTn┬а=arnтАУ1
Here, Tn┬а=┬а1/81,a = 27 and
1/81 = 27r8-1
1/(81├Ч27) = r7
r = 1/3
a1┬а=Ar = 27├Ч1/3 = 9
a2┬а=Ar2┬а= 27├Ч1/9 = 3
a3┬а=Ar3┬а= 27├Ч1/27 = 1
a4┬а=Ar4┬а= 27├Ч1/81 = 1/3
a5┬а=Ar5┬а= 27├Ч1/243 = 1/9
a6┬а=Ar6┬а= 27├Ч1/729 = 1/27
тИ┤┬аThe six GMbetween 27 and 1/81 are 9, 3, 1, 1/3, 1/9, 1/27
Question - 2 : - Insert 5 geometric means between 16 and 1/4.
Answer - 2 : -
Let the five terms bea1, a2, a3, a4, a5.
A = 27, B = 1/81
Now,┬аthese 5terms are between A and B.
So the GP is: A, a1,a2, a3, a4, a5, B.
So we now have 7 termsin GP with the first term being 16 and seventh being 1/4.
We know that,┬аTn┬а=arnтАУ1
Here, Tn┬а=┬а1/4,a = 16 and
1/4 = 16r7-1
1/(4├Ч16) = r6
r = 1/2
a1┬а=Ar = 16├Ч1/2 = 8
a2┬а=Ar2┬а= 16├Ч1/4 = 4
a3┬а=Ar3┬а= 16├Ч1/8 = 2
a4┬а=Ar4┬а= 16├Ч1/16 = 1
a5┬а=Ar5┬а= 16├Ч1/32 = 1/2
тИ┤┬аThe five GMbetween 16 and 1/4 are 8, 4, 2, 1, ┬╜
Question - 3 : - Insert 5 geometric means between 32/9 and 81/2.
Answer - 3 : -
Let the five terms bea1, a2, a3, a4, a5.
A = 32/9, B = 81/2
Now,┬аthese 5terms are between A and B.
So the GP is: A, a1,a2, a3, a4, a5, B.
So we now have 7 termsin GP with the first term being 32/9 and seventh being 81/2.
We know that,┬аTn┬а=arnтАУ1
Here, Tn┬а=┬а81/2,a = 32/9 and
81/2 = 32/9r7-1
(81├Ч9)/(2├Ч32) = r6
r = 3/2
a1┬а=Ar = (32/9)├Ч3/2 = 16/3
a2┬а=Ar2┬а= (32/9)├Ч9/4 = 8
a3┬а=Ar3┬а= (32/9)├Ч27/8 = 12
a4┬а=Ar4┬а= (32/9)├Ч81/16 = 18
a5┬а=Ar5┬а= (32/9)├Ч243/32 = 27
тИ┤┬аThe five GMbetween 32/9 and 81/2 are 16/3, 8, 12, 18, 27
Question - 4 : - Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) тАУ8 and тАУ2
Answer - 4 : -
(i)┬а2 and 8
GM between a and b isтИЪab
Let a = 2 and b =8
GM = тИЪ2├Ч8
= тИЪ16
= 4
(ii)┬аa3band ab3
GM between a and b isтИЪab
Let a = a3band b = ab3
GM = тИЪ(a3b├Ч ab3)
= тИЪa4b4
= a2b2
(iii)┬атАУ8 and тАУ2
GM between a and b isтИЪab
Let a = тАУ2 and b = тАУ8
GM = тИЪ(тАУ2├ЧтАУ8)
= тИЪтАУ16
= -4
Question - 5 : - If a is the G.M. of 2 and┬а┬╝ find a.
Answer - 5 : -
We know that GMbetween a and b is тИЪab
Let a = 2 and b = 1/4
GM = тИЪ(2├Ч1/4)
= тИЪ(1/2)
= 1/тИЪ2
тИ┤ value of a is 1/тИЪ2
Question - 6 : - Find the two numbers whose A.M. is 25 and GM is 20.
Answer - 6 : -
Given: A.M = 25, G.M =20.
G.M = тИЪab
A.M = (a+b)/2
So,
тИЪab = 20 тАжтАж. (1)
(a+b)/2 = 25тАжтАж. (2)
a + b = 50
a = 50 тАУ b
Putting the value ofтАШaтАЩ in equation (1), we get,
тИЪ[(50-b)b] = 20
50b тАУ b2┬а=400
b2┬атАУ50b + 400 = 0
b2┬атАУ40b тАУ 10b + 400 = 0
b(b тАУ 40) тАУ 10(b тАУ 40)= 0
b = 40 or b = 10
If b = 40 then a = 10
If b = 10 then a = 40
тИ┤ The numbers are 10and 40.
Question - 7 : - Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Answer - 7 : -
Let the root of thequadratic equation be┬аa┬аand┬аb.
So, according to thegiven condition,
A.M = (a+b)/2 = A
a + b = 2A тАж.. (1)
GM = тИЪab = G
ab = G2тАж(2)
The quadratic equationis given by,
x2┬атАУ┬аx┬а(Sumof roots) + (Product of roots) = 0
x2┬атАУ┬аx┬а(2A)+ (G2) = 0
x2┬атАУ2Ax┬а+ G2┬а= 0 [Using (1) and (2)]
тИ┤ The requiredquadratic equation is┬аx2┬атАУ 2Ax┬а+ G2┬а=0.