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

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Question

(a) Given that $a,\;b,\;c$ are in $A.P.$

$ \Rightarrow $$2b = a + c$ ......$(i)$

and ${a^2},{b^2},\;{c^2}$ are in $H.P.$

$ \Rightarrow $${b^2

Vedclass · Accepted Answer

$a = b = c$

Vedclass · Answer

(a) Given that $a,\;b,\;c$ are in $A.P.$

$ \Rightarrow $$2b = a + c$ ......$(i)$

and ${a^2},{b^2},\;{c^2}$ are in $H.P.$

$ \Rightarrow $${b^2} = \frac{{2{a^2}{c^2}}}{{{a^2} + {c^2}}}$

$ \Rightarrow $ ${b^2}({a^2} + {c^2}) = 2{a^2}{c^2}$

$ \Rightarrow $ ${b^2}({(a+c)^2} -2ac) = 2{a^2}{c^2}$

$ \Rightarrow $ ${b^2}\left\{ {4{b^2} - 2ac} \right\} = 2{a^2}{c^2}$, from $(i)$

$ \Rightarrow $ $4{b^4} - 2ac{b^2} = 2{a^2}{c^2}$

$ \Rightarrow $ $({b^2} - ac)(2{b^2} + ac) = 0$

$ \Rightarrow $ Either ${b^2} - ac = 0$ or $2{b^2} + ac = 0$

If $b$, then ${b^2} = ac$

$ \Rightarrow $ ${\left\{ {\frac{1}{2}(a + c)} \right\}^2} = ac$from $(i)$

$ \Rightarrow $ ${(a + c)^2} = 4ac$

$\Rightarrow $${(a - c)^2} = 0$

Therefore $a = c$ and if $a = c$ then from ${b^2} = ac$,

we get ${b^2} = {a^2}$ or $b = a$.

Thus $a = b = c$.

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

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