If A.M and G.M of two …
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Posted by Vivek Joshi 8 years, 9 months ago
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Neeraj Sharma 8 years, 9 months ago
{tex}let\,two\,numbers\,be\,a\,and\,b{/tex}
{tex}A.M. = {{a + b} \over 2} = 10{/tex}
{tex}a + b = 20 \ldots \ldots \left( 1 \right){/tex}
{tex}G.M. = \sqrt {ab} = 8{/tex}
{tex}ab = 64 \ldots \ldots \left( 2 \right){/tex}
{tex}Now\,{\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab{/tex}
{tex} = {\left( {20} \right)^2} - 4 \times 64{/tex}
{tex} = 400 - 256{/tex}
{tex}{\left( {a - b} \right)^2} = 144{/tex}
{tex}a - b = \pm 12 \ldots \ldots \left( 3 \right){/tex}
{tex}solving\,\left( 1 \right)\,and\,\left( 3 \right){/tex}
{tex}a = 4,b = 16\,or\,a = 16,b = 4{/tex}
{tex}hence\,two\,numbers\,are\,4\,and\,16\,or\,16\,and\,4.{/tex}
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