If a, b, c are in A.P. and a, b, d are in G.P | vedclass

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(B) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $c = 2b - a$. Since $a, b, d$ are in $G.P.$,we have $b^2 = ad$,which implies

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$G.P.$

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(B) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $c = 2b - a$. Since $a, b, d$ are in $G.P.$,we have $b^2 = ad$,which implies $d = \frac{b^2}{a}$. We need to check the sequence $a, a - b, d - c$. Let the terms be $T_1 = a$,$T_2 = a - b$,and $T_3 = d - c$. Substituting $c$ and $d$: $T_3 = \frac{b^2}{a} - (2b - a) = \frac{b^2 - 2ab + a^2}{a} = \frac{(a - b)^2}{a}$. Now,check the ratio $\frac{T_2}{T_1} = \frac{a - b}{a}$ and $\frac{T_3}{T_2} = \frac{(a - b)^2 / a}{a - b} = \frac{a - b}{a}$. Since the ratios are equal,the sequence $a, a - b, d - c$ is in $G.P.$

If $a, b, c$ are in $A.P.$ and $a, b, d$ are in $G.P.$,then $a, a - b, d - c$ will be in

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