Question

If the A.M between two number a and b is 4 times their geometric mean.Prove that the ratio between the number can be written as
a:b=4+15 : 4-15​

Answer 1

Answer:

Step-by-step explanation:

Arithmetic mean of two numbers A and B is (A+B)/2

Geometric mean is $\sqrt{AB}$ .

as arithmetic mean is 4 times the geometric mean , we can say

      (A+B)/2 = 4×( $\sqrt{AB}$ )

=>   A+B = 8 ×  $\sqrt{AB}$ .

squaring on both sides;

=> $(A+B)^{2}$ = 64AB                                                         -  1.

=>   $A^{2}$ + $B^{2}$ + 2AB = 64AB

=>   $A^{2}$ + $B^{2}$ = 62AB

=>   $A^{2}$ + $B^{2}$ - 2AB = 60AB

=>    $(A-B)^{2}$ = 60AB                                                       - 2.

Divide 1 by 2

(A+B)/(A-B) = $\sqrt{64/60}$

$\sqrt{15}$ × (A+B) = 4 × (A-B)

A(4 - $\sqrt{15}$) = B(4 + $\sqrt{15}$)

so A:B = (4+$\sqrt{15}$) / (4 - $\sqrt{15}$).