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If A, B, C Are in G.P., Then Prove That: a 2 + a B + B 2 B C + C a + a B = B + a C + B

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Question

If a, b, c are in G.P., then prove that:

\[\frac{a^2 + ab + b^2}{bc + ca + ab} = \frac{b + a}{c + b}\]
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Solution

\[\text{ a, b and c are in G . P }. \]

\[ \therefore b^2 = ac . . . . . . . . (i)\]

\[\text { Now, LHS }= \frac{a^2 + ab + b^2}{bc + ca + ab}\]

\[ = \frac{a^2 + ab + ac}{bc + b^2 + ab} \left[ \text { Using } (i) \right]\]

\[ = \frac{a\left( a + b + c \right)}{b\left( c + b + a \right)}\]

\[ = \frac{a}{b}\]

\[ = \frac{1}{r}\]

\[\text { Here, r = common ratio }\]

\[\text { RHS }= \frac{b + a}{c + b}\]

\[ = \frac{ar + a}{a r^2 + ar}\]

\[ = \frac{a(r + 1)}{ar(r + 1)}\]

\[ = \frac{1}{r}\]

\[ \therefore\text {  LHS = RHS }\]

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Chapter 20: Geometric Progression - Exercise 20.5 [Page 46]

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R.D. Sharma Mathematics [English] Class 11
Chapter 20 Geometric Progression
Exercise 20.5 | Q 13 | Page 46

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