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

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Question

(c) Given that $a,\;b,\;c$ are in $G.P.$

So, ${b^2} = ac$&hellip;..(i)

$x = \frac{{a + b}}{2}$&hellip;..(ii)

$y = \frac{{b + c}}{2}$&hellip;.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(c) Given that $a,\;b,\;c$ are in $G.P.$

So, ${b^2} = ac$&hellip;..(i)

$x = \frac{{a + b}}{2}$&hellip;..(ii)

$y = \frac{{b + c}}{2}$&hellip;..(iii)

Now $\frac{a}{x} + \frac{c}{y} = \frac{{2a}}{{a + b}} + \frac{{2c}}{{b + c}} = \frac{{2(ab + bc + 2ca)}}{{ab + ac + {b^2} + bc}}$

$ = \frac{{2(ab + bc + 2ca)}}{{(ab + ac + ac + bc)}} = 2$,$\left\{ {\because \;{b^2} = ac} \right\}$.

Trick : Let $a = 1,\;b = 2,\;c = 4,$ then obviously $x = \frac{3}{2}$

and $y = 3$, then $\frac{1}{{3/2}} + \frac{4}{3} = 2$.

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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