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

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Question

(d) $\frac{{{T_2}}}{{{T_1}}} = \frac{{{T_3}}}{{{T_2}}}$ $ \Rightarrow {2^{(b - a)x}} = {2^{(c - b)x}}$

$ \Rightarrow (b - a)x = (c - b)x$

==>

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $\frac{{{T_2}}}{{{T_1}}} = \frac{{{T_3}}}{{{T_2}}}$ $ \Rightarrow {2^{(b - a)x}} = {2^{(c - b)x}}$

$ \Rightarrow (b - a)x = (c - b)x$

==> $(b - a) = (c - b)$ $\forall \,x,x 
e 0$

$	herefore {2^{ax + 1}},{2^{bx + 1}},{2^{cx + 1}}$  is a $G.P.$,  $\forall \,\,x 
e 0.$

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

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