If G is the geometric mean between a and b- show that 1G-a -1G-b-1G-

G=√( ab ) 1 (G+a) + 1 (G+b) G+b+G+a (G+a)(G+b) 2G+a+b G ^ 2 +aG+bG+ab 2√( ab ) +a+b 2ab+a√( ab+ ) b√( ab ) #160; 2√( a ...

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Byju's Answer

Standard IX

Mathematics

Expansion of (a ± b ± c)²

If G is the...

Question

If $G$ is the geometric mean between $a$ and $b$ , show that $\frac{1}{G + a} + \frac{1}{G + b} = \frac{1}{G}$ .

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Solution

$G = \sqrt{a b} \frac{1}{(G + a)} + \frac{1}{(G + b)} \frac{G + b + G + a}{(G + a) (G + b)} \frac{2 G + a + b}{G^{2} + a G + b G + a b} \frac{2 \sqrt{a b} + a + b}{2 a b + a \sqrt{a b +} b \sqrt{a b}} \frac{2 \sqrt{a b} + a + b}{\sqrt{a b} (2 \sqrt{a b} + a + b)} = \frac{1}{\sqrt{a b}} = \frac{1}{G}$

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If G is the geometric mean between a and b, show that 1G+a+1G+b=1G.

G=√ab1(G+a)+1(G+b)G+b+G+a(G+a)(G+b)2G+a+bG2+aG+bG+ab2√ab+a+b2ab+a√ab+b√ab2√ab+a+b√ab(2√ab+a+b)=1√ab=1G

If $G$ is the geometric mean between $a$ and $b$ , show that $\frac{1}{G + a} + \frac{1}{G + b} = \frac{1}{G}$ .