If a, b, c are in A.P. and a^2, b^2, c^2 are | vedclass

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(D) Given $a, b, c$ are in $A.P.$,we have $2b = a + c$. Since $a^2, b^2, c^2$ are in $H.P.$,we have $\frac{1}{b^2} - \frac{1}{a^2} = \frac{1}{c^2} - \

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$\frac{-a}{2}, b, c$ are in $G.P.$

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(D) Given $a, b, c$ are in $A.P.$,we have $2b = a + c$. Since $a^2, b^2, c^2$ are in $H.P.$,we have $\frac{1}{b^2} - \frac{1}{a^2} = \frac{1}{c^2} - \frac{1}{b^2}$. This simplifies to $\frac{a^2 - b^2}{a^2 b^2} = \frac{b^2 - c^2}{b^2 c^2}$,which implies $\frac{(a-b)(a+b)}{a^2} = \frac{(b-c)(b+c)}{c^2}$. Since $a-b = b-c$,if $a 
eq b$,we can divide by $(a-b)$ to get $\frac{a+b}{a^2} = \frac{b+c}{c^2}$. $c^2(a+b) = a^2(b+c) \Rightarrow c^2 a + c^2 b = a^2 b + a^2 c$. $ac(c-a) = b(a^2-c^2) = b(a-c)(a+c)$. Since $a 
eq c$,we divide by $(c-a)$ to get $ac = -b(a+c)$. Substituting $a+c = 2b$,we get $ac = -b(2b) = -2b^2$,or $b^2 = \frac{-ac}{2} = (\frac{-a}{2})c$. This confirms that $\frac{-a}{2}, b, c$ are in $G.P.$

If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $H.P.$,then

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