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, these 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, Tn =arn–1
Here, Tn = 1/81,a = 27 and
1/81 = 27r8-1
1/(81×27) = r7
r = 1/3
a1 =Ar = 27×1/3 = 9
a2 =Ar2 = 27×1/9 = 3
a3 =Ar3 = 27×1/27 = 1
a4 =Ar4 = 27×1/81 = 1/3
a5 =Ar5 = 27×1/243 = 1/9
a6 =Ar6 = 27×1/729 = 1/27
∴ The 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, these 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, Tn =arn–1
Here, Tn = 1/4,a = 16 and
1/4 = 16r7-1
1/(4×16) = r6
r = 1/2
a1 =Ar = 16×1/2 = 8
a2 =Ar2 = 16×1/4 = 4
a3 =Ar3 = 16×1/8 = 2
a4 =Ar4 = 16×1/16 = 1
a5 =Ar5 = 16×1/32 = 1/2
∴ The five GMbetween 16 and 1/4 are 8, 4, 2, 1, ½
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, these 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, Tn =arn–1
Here, Tn = 81/2,a = 32/9 and
81/2 = 32/9r7-1
(81×9)/(2×32) = r6
r = 3/2
a1 =Ar = (32/9)×3/2 = 16/3
a2 =Ar2 = (32/9)×9/4 = 8
a3 =Ar3 = (32/9)×27/8 = 12
a4 =Ar4 = (32/9)×81/16 = 18
a5 =Ar5 = (32/9)×243/32 = 27
∴ The 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) –8 and –2
Answer - 4 : -
(i) 2 and 8
GM between a and b is√ab
Let a = 2 and b =8
GM = √2×8
= √16
= 4
(ii) a3band ab3
GM between a and b is√ab
Let a = a3band b = ab3
GM = √(a3b× ab3)
= √a4b4
= a2b2
(iii) –8 and –2
GM between a and b is√ab
Let a = –2 and b = –8
GM = √(–2×–8)
= √–16
= -4
Question - 5 : - If a is the G.M. of 2 and ¼ find a.
Answer - 5 : -
We know that GMbetween a and b is √ab
Let a = 2 and b = 1/4
GM = √(2×1/4)
= √(1/2)
= 1/√2
∴ value of a is 1/√2
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 = √ab
A.M = (a+b)/2
So,
√ab = 20 ……. (1)
(a+b)/2 = 25……. (2)
a + b = 50
a = 50 – b
Putting the value of‘a’ in equation (1), we get,
√[(50-b)b] = 20
50b – b2 =400
b2 –50b + 400 = 0
b2 –40b – 10b + 400 = 0
b(b – 40) – 10(b – 40)= 0
b = 40 or b = 10
If b = 40 then a = 10
If b = 10 then a = 40
∴ 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 a and b.
So, according to thegiven condition,
A.M = (a+b)/2 = A
a + b = 2A ….. (1)
GM = √ab = G
ab = G2…(2)
The quadratic equationis given by,
x2 – x (Sumof roots) + (Product of roots) = 0
x2 – x (2A)+ (G2) = 0
x2 –2Ax + G2 = 0 [Using (1) and (2)]
∴ The requiredquadratic equation is x2 – 2Ax + G2 =0.