(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