If a- b and c are in G-P- then b-ab-c-b-ab-c-

The correct option is B 0 If a,b,c are in G.P, then b^2=ac Or #160; b=√(ac) Consider, b ab c+b+ab+c =√(ac) a√(ac) c+√(ac)+a√(ac)+c ...

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If a, b and c...

Question

If a, b and c are in G.P., then $\frac{b - a}{b - c} + \frac{b + a}{b + c} =$

b^{2}

c^{2}

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a c

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a b

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0

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Similar questions

\frac{a b - c d}{b^{2} - c^{2}} = \frac{a + c}{b}

a, b, c, d

G . P

\frac{a b - c d}{b^{2} - c^{2}} = \frac{a + c}{b}

a^{2} + b^{2}, a b + b c

b^{2} + c^{2}

a, b, c

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

(i) $\frac{a b - c d}{b^{2} - c^{2}} = \frac{a + c}{b}$

(ii) $(a + b + c + d)^{2} = (a + b)^{2} + 2 (b + c)^{2} + (c + d)^{2}$

(iii) $(b + c) (b + d) = (c + d) (c + d)$

\frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b} = 3

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