If a, b, c are in A.P. as well as in G.P.,the | vedclass

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(D) Given that $a, b, c$ are in $A.P.$,we have $b = \frac{a + c}{2} \implies 2b = a + c$ .....$(i)$Given that $a, b, c$ are in $G.P.$,we have $b^2 = a

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$a = b = c$

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(D) Given that $a, b, c$ are in $A.P.$,we have $b = \frac{a + c}{2} \implies 2b = a + c$ .....$(i)$Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$ .....(ii)From $(i)$,$a + c = 2b$. Squaring both sides,we get $(a + c)^2 = 4b^2$.Substituting $b^2 = ac$ from (ii) into the equation,we get $(a + c)^2 = 4ac$.This simplifies to $a^2 + 2ac + c^2 = 4ac$,which is $a^2 - 2ac + c^2 = 0$.Thus,$(a - c)^2 = 0$,which implies $a = c$.Substituting $a = c$ into $(i)$,we get $b = \frac{a + a}{2} = a$.Therefore,$a = b = c$.

If $a, b, c$ are in $A.P.$ as well as in $G.P.$,then

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