I don't know why I do this, but here is the general case. Suppose $am = gm+u$ and $gm = hm+v$ with $u, v \ne 0$ and $u \ne v$. Since $hm \le gm \le am $, $u \ge 0$ and $v \ge 0$. Since $gm^2 = am\cdot hm$, $gm^2 =(gm+u)(gm-v) =gm^2+gm(u-v)-uv $ so $gm(u-v) =uv$. Therefore $u > v$ and $gm =\dfrac{uv}{u-v} $. Then $am =gm+u =\dfrac{uv}{u-v}+u =\dfrac{uv+u(u-v)}{u-v}+u =\dfrac{u^2}{u-v} $ and $hm =gm-v =\dfrac{uv}{u-v}-v =\dfrac{uv-v(u-v)}{u-v} =\dfrac{v^2}{u-v} $. If there are only two values, $a$ and $b$, then $\dfrac{a+b}{2} =\dfrac{u^2}{u-v} $ and $\sqrt{ab} =\dfrac{uv}{u-v} $ so $b =\dfrac{u^2v^2}{a(u-v)^2} $ and $\dfrac{u^2}{u-v} =\dfrac{a+\dfrac{u^2v^2}{a(u-v)^2}}{2} =\dfrac{a^2(u-v)^2+u^2v^2}{2a(u-v)^2} $ or $2au^2(u-v) =a^2(u-v)^2+u^2v^2 $. Solving $a^2(u-v)^2-2u^2(u-v)a+u^2v^2 =0$, $\begin{array}\\ a &=\dfrac{2u^2(u-v)\pm\sqrt{4u^4(u-v)^2-4(u-v)^2u^2v^2}}{2(u-v)^2}\\ &=\dfrac{u^2(u-v)\pm u(u-v)\sqrt{u^2-v^2}}{(u-v)^2}\\ &=\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\ &=u\dfrac{u\pm \sqrt{u^2-v^2}}{u-v}\\ \text{and}\\ b &=\dfrac{2u^2}{u-v}-a\\ &=\dfrac{2u^2}{u-v}-\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\ &=\dfrac{2u^2-u^2\mp u\sqrt{u^2-v^2}}{u-v}\\ &=\dfrac{u^2\mp u\sqrt{u^2-v^2}}{u-v}\\ \end{array} $ For this case, $u=2$ and $v=1.6$. $gm =\dfrac{2\cdot 1.6}{2-1.6} =\dfrac{3.2}{.4} =8 $. To get $a$ and $b$, $\sqrt{u^2-v^2} =\sqrt{4-2.56} =\sqrt{1.44} =1.2 $ so $a =2\dfrac{2\pm 1.2}{.4} =2\dfrac{3.2, .8}{.4} =2(8, 2) =(16, 4) $ and $b =(4, 16) $.
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Question: Can anyone tell me how to solve this problem? I do not know how since from what I've known, the AM of is always greater than GM. Please show me your complete solution $\endgroup$ 3 > 1
120 Qs. 480 Marks 120 Mins
Concept: Let x and y be the two numbers. The the arithmetic mean A, geometric mean G and the harmonic mean H of x and y is given by, ⇒ A = ⇒ G2 = xy ⇒ Calculations: Consider, the two numbers are x and y. Given, the arithmetic mean and geometric mean of the x and y is A and G. ⇒ A = ....(1) ⇒ G2 = xy ....(2) The harmonic mean of two number x and y is 4. ⇒ ⇒ 2xy = 4(x + y) ⇒ ⇒ G2 = 4A (∵ x + y = 2A) ⇒ G2 = 4A ....(3) Given, Their arithmetic mean A and the geometric mean G satisfy the relation 2A + G2 = 27. ⇒2A + G2 = 27 ⇒ 6A = 27 ⇒ A = From equation (1), (2) and (3), we have x + y = 9 and xy = 18 ⇒ x = 6 and y = 3 Hence, the harmonic mean of two number is 4, Their arithmetic mean A and the geometric mean G satisfy the relation 2A + G2 = 27, then the two numbers are 6 and 3. India’s #1 Learning Platform Start Complete Exam Preparation
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The arithmetic mean of two numbers is 4 and their harmonic mean is 15/4. find the numbers
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