A-beya-bey - b-ceyb-cey - c-deyc-dey- then a-b-c-d are in

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Basic Differentiation Rule

a+beya-bey = ...

Question

$\frac{a + b e^{y}}{a - b e^{y}} = \frac{b + c e^{y}}{b - c e^{y}} = \frac{c + d e^{y}}{c - d e^{y}}$ , then $a, b, c, d$ are in

A . P .

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G . P .

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H . P .

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A . G . P .

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Solution

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Similar questions

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

a, b, c, d

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

a, b, c

b, c, d

c, d, e

a, c, e

e = \frac{(b - a)^{2}}{a}

a, b, c

H . P

b, c, d

G . P

c, d, e

A . P

\frac{a b^{2}}{(2 a - b)^{2}}

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