## 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 = \(\sqrt { AH }\)

⇒ G^{2} = (G + 2) (G – 1.2)

⇒ G^{2} = G^{2} + 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.

∴ b^{2} = 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.

A_{1} A_{2} are two A.M. between two positive numbers a and b, G_{1}, G_{2} are two GM. and H_{1}, H_{2} are two H.M. then prove that:

(i) A_{1}H_{2} = A_{2}H_{1} = G_{1} G_{2} = ab

(ii) G_{1}G_{2} : H_{1}H_{2} = (A_{1} + A_{2}): (H_{1} + H_{2})

Solution:

Let two numbers are a and b.

Then A_{1}, A_{2} are two A.M. between a and b.

Again, G_{1}, G_{2} are two GM. between a and b.

Then, a, G_{1}, G_{2}, 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.

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