a, b,c, d∈R+ such that a, b, c are in A.P. and b, c, d are inH.P. then :

b=12(a+c), c=2bdb+d ⇒ bc+dc=a+cd=ad+dc⇒ bc=ad

a, b,c, d∈R+ such that a, b, c are in A.P. and b, c, d are inH.P. then :

b=12(a+c), c=2bdb+d ⇒ bc+dc=a+cd=ad+dc⇒ bc=ad

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$a, b, c, d \in R^{+}$ such that $a, b, c$ are in A.P. and $b, c, d$ are in $H . P .$ then :

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$a b = c d$

$a c = b d$

$b c = a d$

None

answer is C.

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$b = \frac{1}{2} (a + c), c = \frac{2 b d}{b + d}$

$\Rightarrow$ $b c + d c = (a + c) d = a d + d c$

$\Rightarrow$ $b c = a d$

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a, b,c, d∈R+ such that a, b, c are in A.P. and b, c, d are inH.P. then :

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$a, b, c, d \in R^{+}$ such that $a, b, c$ are in A.P. and $b, c, d$ are in $H . P .$ then :