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8. Sequences and Series
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If $a,b,c$ are in $A.P.$, then ${2^{ax + 1}},{2^{bx + 1}},\,{2^{cx + 1}},x \ne 0$ are in
A
$A.P.$
B
$G.P.$ only when $x > {\rm{0}}$
C
$G.P.$ if $x < 0$
D
$G.P.$ for all $x \ne 0$
Solution
(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$
==> $(b – a) = (c – b)$ $\forall \,x,x \ne 0$
$\therefore {2^{ax + 1}},{2^{bx + 1}},{2^{cx + 1}}$ is a $G.P.$, $\forall \,\,x \ne 0.$
Standard 11
Mathematics
