RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.8

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.8

Question 1.
A.M. of two numbers is 60 and H.M. is 18, Find the numbers.
Solution:
Let two numbers are a and b, then
\(\frac { a+b }{ 2 } \) = 50 ⇒ a + b = 100
and \(\frac { 2ab }{ a+b } \) = 18
From equation (i) and (ii)
\(\frac { 2ab }{ 100 } \) = 18 ⇒ ab = 900
Now (a – b)² = (a + b)² – 4ab
= (100)² – 4 × 900
= 10000 – 3600 = 6400
a – b = \(\sqrt { 6400 }\) = 80
Now, solving a + b = 100 and a – b = 80.
We get a = 90 and b = 10
Hence, numbers are 90 and 10.

Question 2.
If ratio of H.M. and GM. of two numbers is
12 : 13, then prove that ratio of number is 4 : 9.
Solution:
Let two number are a and 6 then its

Question 3.
The difference between A.M. and GM. is 2, difference between GM. and H.M., is 1.2. Find the numbers.
Solution:
According to question
A – G = 2 and A = G + 2
and G – H = 1.2 and H = G – 1.2
We know that,
G = \(\sqrt { AH }\)
⇒ G² = (G + 2) (G – 1.2)
⇒ G² = G² + 2G – 1.2G – 2.4
⇒ 0.8G = 2.4
⇒ G = 3
∴ A = 5, G = 3, H = 1.8

∴ a – b = 8 …(iii)
Solving equation (i) and (iii)
a = 9 and b = 1
Hence, number are 9 and 1.

Question 4.
Three numbers a, b, c are in GP. and ax = by = c^z, then prove that x, z will be in H.P.
Solution:
Let a^x = b^y = c^z = k
Then a = k^1/x, b = k^1/y, c = k^1/2 and a, b, c are in GP.
∴ b² = ac …(i)
Put the value of a, b, c in equation (i)
k^1/y = k^1/x, k^1/z or k^2/y = k^1/x+1/z
which is only possible, when
\(\frac { 2 }{ y } \) = \(\frac { 1 }{ x } \) + \(\frac { 1 }{ z } \)
This, shows that \(\frac { 1 }{ x } \),\(\frac { 1 }{ y } \),\(\frac { 1 }{ z } \) are in A.P. Then x, y, z will be in H.P.

Question 5.
Three numbers a, b, c are in H.P. Prove that 2a – b, b, 2c – b will be in GP.
Solution:

It is clear that from equation (i) and (ii), if a, b, c are in H.P. then 2a – b, b, 2c – b will be in G.P.

Question 6.
If a, b, c are in A.P., x,y, z are in H.P. and ax, by, cz are in GP., then prove that

Solution:

Question 7.
A₁ A₂ are two A.M. between two positive numbers a and b, G₁, G₂ are two GM. and H₁, H₂ are two H.M. then prove that:
(i) A₁H₂ = A₂H₁ = G₁ G₂ = ab
(ii) G₁G₂ : H₁H₂ = (A₁ + A₂): (H₁ + H₂)
Solution:
Let two numbers are a and b.
Then A₁, A₂ are two A.M. between a and b.

Again, G₁, G₂ are two GM. between a and b.
Then, a, G₁, G₂, b will be in G.P.

Question 8.
If a, b, c are in A.P., b, c, d are in GP. and c, d, e are in H.P., then prove that a, c, e will be in H.P.
Solution:
∵ a, b, c are in A.P.

Put the value of b and d from (i) and (iii) in equation (ii),

Question 9.
If three numbers a, b, c are in A.P. and H.P.
both than prove that they will be in GP. also.
Solution:
a, b, c will be in A.P. and H.P.

RBSE Solutions for Class 11 Maths