Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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If A, B, C, D Are in G.P., Prove That: a B − C D B 2 − C 2 = a + C B - Mathematics

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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Exercise 20.5 | Q 9.1 | Page 46
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