Find two positive numbers a and b, whose

Solution:

(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 $b=\sqrt{a b}$

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

$\frac{a+b}{2}=25$

$\Rightarrow a+b=50$

$\Rightarrow b=50-a \ldots$ (i)

Geometric mean of two numbers $=\sqrt{a b}$

$\Rightarrow \sqrt{a b}=20$

$\Rightarrow a b=400$

Substituting value of b from eqn. (i)

$a(50-a)=400$

$\Rightarrow 50 a-a^{2}=400$

On rearranging

$\Rightarrow a^{2}-50 a+400=0$

$\Rightarrow a^{2}-40 a-10 a+400$

$\Rightarrow a(a-40)-10(a-40)=0$

$\Rightarrow(a-10)(a-40)=0$

$\Rightarrow a=10,40$

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 $b=\sqrt{a b}$

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

$\frac{a+b}{2}=10$

$\Rightarrow a+b=20$

$\Rightarrow a=20-b \ldots$ (i)

Geometric mean of two numbers $=\sqrt{a b}$

$\Rightarrow \sqrt{a b}=8$

$\Rightarrow a b=64$

Substituting value of a from eqn. (i)

$b(20-b)=64$

$\Rightarrow 20 b-b^{2}=64$

On rearranging

$\Rightarrow b^{2}-20 b+64=0$

$\Rightarrow b^{2}-16 b-4 b+64$

$\Rightarrow b(b-16)-4(b-16)=0$

$\Rightarrow(b-16)(b-4)=0$

$\Rightarrow b=16,4$

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

a = 4 Or b = 16

Therefore two numbers are 16 and 4