Question

The A.M of two positive numbers whose G.M and H.M are 4 and 3.2 is

Answer 1

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{first \: + ve \: number \: be \: x} \\ &\sf{second \: + ve \: number \: is \: y} \end{cases}\end{gathered}\end{gathered}$

Given that,

• Geometric mean, G = 4

and

• Harmonic mean, H = 3.2

We know,

☆ Arithmetic mean (A) between two positive numbers x and y is given by

$\bf :\longmapsto\:A = \dfrac{x + y}{2} - - - (1)$

☆ Geometric mean (G) between two positive numbers x and y is given by

$\bf :\longmapsto\:G = \sqrt{xy}$

$\bf\implies \: {G}^{2} = xy - - - (2)$

☆ Harmonic mean (H) between two positive numbers x and y is given by

$\rm :\longmapsto\:H = \dfrac{2xy}{x + y}$

$\rm :\longmapsto\:H = \dfrac{2}{x + y} \times xy$

$\rm :\longmapsto\:H = \dfrac{1}{A} \times {G}^{2}$

$\: \: \: \: \: \: \sf \: \red{\bigg \{ \because \: using \: equation \: (1) \: and \: (2)\bigg \}}$

$\bf\implies \:A = \dfrac{ {G}^{2} }{H}$

$\rm :\longmapsto\:A = \dfrac{ {4}^{2} }{3.2}$

$\rm :\longmapsto\:A = \dfrac{16 \times 10}{32}$

$\bf\implies \:A = 5$

Hence,

• Arithmetic mean between two positive numbers = 5.

Additional Information :-

☆ If A, G and H are Arithmetic mean, Geometric mean and Harmonic mean between n numbers, then

$\rm :\longmapsto\:A \geqslant G \geqslant H$

$\rm :\longmapsto\: {G}^{2} = AH$

☆ Arithmetic mean and Geometric mean between two positive numbers satisfy the following quadratic

$\rm :\longmapsto\: {x}^{2} - 2Ax + {G}^{2} = 0$