If a,b,c are in AP, then 10ax+10,10bx+10,10cx+10 (x≠0) are in

Since a, b, c care in AP. ∴ 2b =a+c⇒ 2bx =(a+c)x for all x⇒ 2bx+20 =(a+c)x+20 for all x⇒ 2(bx+10) =(ax+10)+(cx+10) ∴ 102(bx+10) =10(ax+10)+(cx+10)⇒ 10bx+102 =10ax+10⋅10cx+10⇒ 10ax+10, 10bx+10,10cx+10 are in GP for all x

If a,b,c are in AP, then 10ax+10,10bx+10,10cx+10 (x≠0) are in

Since a, b, c care in AP. ∴ 2b =a+c⇒ 2bx =(a+c)x for all x⇒ 2bx+20 =(a+c)x+20 for all x⇒ 2(bx+10) =(ax+10)+(cx+10) ∴ 102(bx+10) =10(ax+10)+(cx+10)⇒ 10bx+102 =10ax+10⋅10cx+10⇒ 10ax+10, 10bx+10,10cx+10 are in GP for all x

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If $a, b, c$ are in AP, then $10^{ax + 10}, 10^{bx + 10}, 10^{cx + 10}$ $(x \neq 0)$ are in

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GP only when $x < 0$

GP for all $x$

GP only when $x > 0$

answer is C.

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Detailed Solution

Since a, b, c care in AP.

$\begin{array}{lrl} ∴ 2 b = a + c \\ \Rightarrow 2 bx = (a + c) x for all x \\ \Rightarrow 2 bx + 20 = (a + c) x + 20 for all x \\ \Rightarrow 2 (bx + 10) = (ax + 10) + (cx + 10) \\ ∴ 10^{2 (bx + 10)} = 10^{(ax + 10) + (cx + 10)} \\ \Rightarrow {(10^{bx + 10})}^{2} = 10^{ax + 10} \cdot 10^{cx + 10} \\ \Rightarrow 10^{ax + 10}, 10^{bx + 10}, 10^{cx + 10} are in GP for all x \end{array}$