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Question
If a, b, c, d are in G.P., prove that:
\[\frac{ab - cd}{b^2 - c^2} = \frac{a + c}{b}\]
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Solution
a, b, c and d are in G.P.
\[\therefore b^2 = ac\]
\[bc = ad\]
\[ c^2 = bd\] .......(1)
\[\text { LHS } = \frac{ab - cd}{b^2 - c^2}\]
\[ = \frac{ab - cd}{ac - bd} \left[\text { Using } (1) \right]\]
\[ = \frac{\left( ab - cd \right)b}{\left( ac - bd \right)b} \]
\[ = \frac{a b^2 - bcd}{\left( ac - bd \right)b}\]
\[ = \frac{a\left( ac \right) - c\left( c^2 \right)}{\left( ac - bd \right)b} \left[ \text { Using } (1) \right]\]
\[ = \frac{a^2 c - c^3}{\left( ac - bd \right)b}\]
\[ = \frac{c\left( a^2 - c^2 \right)}{\left( ac - bd \right)b}\]
\[ = \frac{\left( a + c \right)\left( ac - c^2 \right)}{\left( ac - bd \right)b}\]
\[ = \frac{\left( a + c \right)\left( ac - bd \right)}{\left( ac - bd \right)b} \left[\text{ Using } (1) \right]\]
\[ = \frac{\left( a + c \right)}{b} = \text { RHS }\]
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