Question

If the arithmetic mean and the geometric mean if two numbers are 34 and 16, find the numbers.​

Answer 1

Let the two numbers be 'a' and 'b'.

We know,

♠ Arithmetic mean(A.M.) = (a + b)/2 = 34 (given)

♠ Also, Geometric mean(G.M.) = √ab = 16 (given)

We get :

(a + b) = 2*34 = 68           ...(i)

ab = 16² = 256

Now,

(a - b)² = (a + b)² - 4ab

⇒(a - b)² = (68)² - 4*256

⇒(a - b)² = 4624 - 1024

⇒(a - b)² = 3600

⇒(a - b) = 60                    ...(ii)

From eq. (ii) :

a = 60 + b                         ...(iii)

Putting this value in eq. (i) :

(a + b) = 68

⇒60 + b + b = 68

⇒60 + 2b = 68

⇒2b = 8

$\boxed{\implies b = 4}$

Putting b = 4 in eq. (iii) :

a = 60 + b

⇒a = 60 + 4

$\boxed{\implies a = 64}$

∴ So, the numbers are 64 and 4.

Answer 2

Answer :

The numbers are either 64 and 4

or 4 and 64

Given :

• The arithmetic mean and geometric mean of two numbers are 34 and 16 .

To Find :

• The numbers whose AM and GM are 34 and 16 respectively.

Solution :

Let a and b be the required numbers

$\sf{AM = 34 }\\ \implies \sf\dfrac{a + b}{2} = 34 \\ \sf \implies a + b = 68 \: \: ...........(1)$

and

$\sf{GM = 16} \\ \implies \sf{ \sqrt{ab} } = 16 \\ \implies \sf{ab = {16}^{2} } \\ \implies \sf{ab} = 256 \\ \implies \sf{a = \frac{256}{b} \: \: .............(2)}$

Putting the value of a from (2) in (1) we have :

$\sf \implies \dfrac{256}{b} + b = 68 \\ \implies \sf \dfrac{256 + {b}^{2} }{b} = 68 \\ \implies \sf256 + {b}^{2} = 68b \\ \sf\implies {b}^{2} - 68b + 256 = 0 \\ \sf \implies {b}^{2} - 4b - 64b + 256 = 0 \\ \sf \implies b(b - 4) - 64(b - 4) = 0 \\ \sf\implies(b - 4)(b - 64) = 0$

Now ,

$\sf{b - 4 = 0 \: \: and \: \: b - 64 = 0 } \\ \sf\implies b = 4 \: \: and \: \implies b = 64$

Thus value of b is either 4 or 64

Using the value of b in (2)

$\sf a = \dfrac{256}{4} \: \: and \: \: \: a = \dfrac{256}{64} \\ \sf \implies a = 64 \: \: and \: \: \implies a = 4$

The value of a is either 64 or 4

Therefore , the numbers are either (64 , 4) or (4,64)