If a, b, c are in A.P.,then 2^{ax + 1}, 2^{bx | vedclass

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(D) Given that $a, b, c$ are in $A.P.$,we have $b - a = c - b$.Let the terms be $T_1 = 2^{ax + 1}$,$T_2 = 2^{bx + 1}$,and $T_3 = 2^{cx + 1}$.For these

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$G.P.$ for all $x 
e 0$

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(D) Given that $a, b, c$ are in $A.P.$,we have $b - a = c - b$.Let the terms be $T_1 = 2^{ax + 1}$,$T_2 = 2^{bx + 1}$,and $T_3 = 2^{cx + 1}$.For these terms to be in $G.P.$,the ratio $\frac{T_2}{T_1}$ must equal $\frac{T_3}{T_2}$.$\frac{T_2}{T_1} = \frac{2^{bx + 1}}{2^{ax + 1}} = 2^{(b - a)x}$.$\frac{T_3}{T_2} = \frac{2^{cx + 1}}{2^{bx + 1}} = 2^{(c - b)x}$.Since $b - a = c - b$,it follows that $2^{(b - a)x} = 2^{(c - b)x}$ for all $x 
e 0$.Thus,the terms form a $G.P.$ for all $x 
e 0$.

If $a, b, c$ are in $A.P.$,then $2^{ax + 1}, 2^{bx + 1}, 2^{cx + 1}$ for $x \ne 0$ are in

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