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If $\mathrm{x}, \frac{3}{2}, \mathrm{z}$ are in $\mathrm{AP} ; \mathrm{x}, 3, \mathrm{z}$ are in GP; then which one of the following will be in HP?
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Verified Answer
The correct answer is:
$\mathrm{x}, 6, \mathrm{z}$
$\mathrm{x}, \frac{3}{2}, \mathrm{z}$ are in A.P
If $a, b, c$ are in $A \cdot P 2 b=a+c$
$\therefore 2\left(\frac{3}{2}\right)=x+z$
$\Rightarrow 3=x+z$
$\ldots .(1)$
$\mathrm{x}, 3, \mathrm{z}$ are in G.P. If $a, b, c$ are in G.P. $b^{2}=a c$ $\therefore 3^{2}=\mathrm{xz}$
$\Rightarrow 9=\mathrm{xz}$...(2
If $\mathrm{x}, 6, \mathrm{z}$ are in $\mathrm{H.P.} \frac{2}{6}=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{z}}$
$\quad\left(\because \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\right.$, if $a, b, c$ are is H.P. $)$
$\Rightarrow \frac{1}{3}=\frac{z+x}{x z}=\frac{3}{9}=\frac{1}{3} \quad($ from $(1)(2))$
L.HS. $=$ R.H.S.
If $a, b, c$ are in $A \cdot P 2 b=a+c$
$\therefore 2\left(\frac{3}{2}\right)=x+z$
$\Rightarrow 3=x+z$
$\ldots .(1)$
$\mathrm{x}, 3, \mathrm{z}$ are in G.P. If $a, b, c$ are in G.P. $b^{2}=a c$ $\therefore 3^{2}=\mathrm{xz}$
$\Rightarrow 9=\mathrm{xz}$...(2
If $\mathrm{x}, 6, \mathrm{z}$ are in $\mathrm{H.P.} \frac{2}{6}=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{z}}$
$\quad\left(\because \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\right.$, if $a, b, c$ are is H.P. $)$
$\Rightarrow \frac{1}{3}=\frac{z+x}{x z}=\frac{3}{9}=\frac{1}{3} \quad($ from $(1)(2))$
L.HS. $=$ R.H.S.
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