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If $a, b, c, d$ are in H.P., then $a b+b c+c d$ is equal to
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Verified Answer
The correct answer is:
$3 a d$
Since $a, b, c, d$ are in H.P., therefore $b$ is the H.M. of $a$ and $c$ i.e. $b=\frac{2 a c}{a+c}$ and $c$ is the H.M. of $b$ and
d i.e. $c=\frac{2 b d}{b+d}$ $\therefore \quad(a+c)(b+d)=\frac{2 a c}{b} \cdot \frac{2 b d}{c}$
$\Rightarrow a b+a d+b c+c d=4 a d \Rightarrow a b+b c+c d=3 a d$.
Trick: Check for $a=1, b=\frac{1}{2}$, $c=\frac{1}{3}, d=\frac{1}{4}$
d i.e. $c=\frac{2 b d}{b+d}$ $\therefore \quad(a+c)(b+d)=\frac{2 a c}{b} \cdot \frac{2 b d}{c}$
$\Rightarrow a b+a d+b c+c d=4 a d \Rightarrow a b+b c+c d=3 a d$.
Trick: Check for $a=1, b=\frac{1}{2}$, $c=\frac{1}{3}, d=\frac{1}{4}$
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