If a, b, c are in AP,and b-a, c-b, a are in G | vedclass

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(A) Given that $a, b, c$ are in $AP$,so $2b = a + c$. Since $b-a, c-b, a$ are in $GP$,we have $(c-b)^2 = (b-a)a$. In an $AP$,$b-a = c-b = d$ (common d

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$1: 2: 3$

Vedclass · Answer

(A) Given that $a, b, c$ are in $AP$,so $2b = a + c$. Since $b-a, c-b, a$ are in $GP$,we have $(c-b)^2 = (b-a)a$. In an $AP$,$b-a = c-b = d$ (common difference). Substituting $c-b = b-a$ into the $GP$ condition: $(b-a)^2 = (b-a)a$. Assuming $b 
eq a$,we get $b-a = a$,which implies $b = 2a$. Substituting $b = 2a$ into $2b = a + c$,we get $2(2a) = a + c$,so $4a = a + c$,which implies $c = 3a$. Thus,$a: b: c = a: 2a: 3a = 1: 2: 3$.

If $a, b, c$ are in $AP$,and $b-a, c-b, a$ are in $GP$,then $a: b: c$ is

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