If a, b, c are in G.P. and x, y are the arith | vedclass

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(C) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$ .....$(i)$$x$ is the arithmetic mean between $a$ and $b$,so $x = \frac{a + b}{2}$ .....(ii)$

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$2$

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(C) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$ .....$(i)$$x$ is the arithmetic mean between $a$ and $b$,so $x = \frac{a + b}{2}$ .....(ii)$y$ is the arithmetic mean between $b$ and $c$,so $y = \frac{b + c}{2}$ .....(iii)Now,consider the expression $\frac{a}{x} + \frac{c}{y}$.Substituting the values of $x$ and $y$: $\frac{a}{\frac{a+b}{2}} + \frac{c}{\frac{b+c}{2}} = \frac{2a}{a+b} + \frac{2c}{b+c}$$= \frac{2a(b+c) + 2c(a+b)}{(a+b)(b+c)} = \frac{2ab + 2ac + 2ac + 2bc}{ab + ac + b^2 + bc}$Since $b^2 = ac$,the denominator becomes $ab + ac + ac + bc = ab + 2ac + bc$.Thus,the expression is $\frac{2(ab + 2ac + bc)}{ab + 2ac + bc} = 2$.

If $a, b, c$ are in $G.P.$ and $x, y$ are the arithmetic means between $a, b$ and $b, c$ respectively,then $\frac{a}{x} + \frac{c}{y}$ is equal to

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