If a, b, c are in A.P. and a, c - b, b - a ar | vedclass

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(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. Also,$a, c - b, b - a$ are in $G.P.$,so $(c - b)^2 = a(b - a)$. Since $a, b, c$ are in $A

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$1:2:3$

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(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. Also,$a, c - b, b - a$ are in $G.P.$,so $(c - b)^2 = a(b - a)$. Since $a, b, c$ are in $A.P.$,let $a = x - d, b = x, c = x + d$. Then $c - b = d$ and $b - a = d$. The $G.P.$ condition becomes $d^2 = (x - d)(d)$. Since $a 
e b 
e c$,$d 
e 0$,so $d = x - d$,which implies $x = 2d$. Thus,$a = 2d - d = d$,$b = 2d$,and $c = 2d + d = 3d$. Therefore,$a:b:c = d:2d:3d = 1:2:3$.

If $a, b, c$ are in $A.P.$ and $a, c - b, b - a$ are in $G.P.$ $(a \ne b \ne c)$,then $a:b:c$ is

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