JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 5)
If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric mean, then $${{a + b} \over {a - b}}$$ is equal to :
$${{\sqrt 6 } \over 2}$$
$${{3\sqrt 2 } \over 4}$$
$${{7\sqrt 3 } \over {12}}$$
$${{5\sqrt 6 } \over {12}}$$
Explanation
A.T.Q.,
A.M. = 5G.M.
$${{a + b} \over 2} = 5\sqrt {ab} $$
$${{a + b} \over {\sqrt {ab} }}$$ $$ = 10$$
$$ \therefore $$ $${a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}$$
Use componendo and Dividendo
$${{a + b} \over {a - b}} = {{20} \over {8\sqrt 6 }} = {5 \over {2\sqrt 6 }} = {{5\sqrt 6 } \over {12}}$$
A.M. = 5G.M.
$${{a + b} \over 2} = 5\sqrt {ab} $$
$${{a + b} \over {\sqrt {ab} }}$$ $$ = 10$$
$$ \therefore $$ $${a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}$$
Use componendo and Dividendo
$${{a + b} \over {a - b}} = {{20} \over {8\sqrt 6 }} = {5 \over {2\sqrt 6 }} = {{5\sqrt 6 } \over {12}}$$
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