(i) AM = 25 and GM = 20 

To find: Two positive numbers a and b 

Given: AM = 25 and GM = 20 

Formula used: 

(i) Arithmetic mean between a and b = \(\frac{a + b }{2}\)

(ii) Geometric mean between a and b = \(\sqrt{ab}\)

Arithmetic mean of two numbers = \(\frac{a+b}{2}\)

\(\frac{a+b}{2}\) = 25

⇒ a + b = 50 

⇒ b = 50 – a … (i)

Geometric mean of two numbers = \(\sqrt{ab}\)

\(\Rightarrow\) \(\sqrt{ab}\) = 20

\(\Rightarrow\) ab = 400

Substituting value of b from eqn. (i)

a(50 – a) = 400 

⇒ 50a – a² = 400 

On rearranging 

⇒ a² – 50a + 400 = 0 

⇒ a² – 40a – 10a + 400 

⇒ a(a – 40) – 10(a – 40) = 0 

⇒ (a – 10) (a – 40) = 0 

⇒ a = 10, 40 Substituting, a = 10 Or a = 40 in eqn. (i) 

b = 40 Or b = 10 

Therefore two numbers are 10 and 40 

(ii) AM = 10 and GM = 8 

To find: Two positive numbers a and b 

Given: AM = 10 and GM = 8 

Formula used: 

(i) Arithmetic mean between a and b = \(\frac{a + b}{2}\)

(ii) Geometric mean between a and b = \(\sqrt{ab}\)

Arithmetic mean of two numbers = \(\frac{a+ b }{2}\)

\(\frac{a+ b }{2}\) = 10

⇒ a + b = 20 

⇒ a = 20 – b … (i)

Geometric mean of two numbers = \(\sqrt{ab}\)

⇒ \(\sqrt{ab}\) = 8

⇒ ab = 64

Substituting value of a from eqn. (i) 

b(20 – b) = 64 

⇒ 20b – b² = 64 

On rearranging 

⇒ b² – 20b + 64 = 0 

⇒ b² – 16b – 4b + 64 

⇒ b(b – 16) – 4(b – 16) = 0 

⇒ (b – 16) (b – 4) = 0 

⇒ b = 16, 4 

Substituting, b = 16 Or b = 4 in eqn. (i)

a = 4 Or b = 16 

Therefore two numbers are 16 and 4