Question

Să se determine a∈R astfel încât numerele 2^(a-1), 2^(-a+2) + 1, 2^(a+1) +1 să fie în progresie aritmetică.

Answer 1

$\displaystyle 2^{a-1},~2^{-a+2}+1,~2^{a+1}+1\\ \\ 2^{-a+2}+1=\frac{2^{a-1}+2^{a+1}+1}{2} \Rightarrow 2(2^{-a+2}+1)=2^{a-1}+2^{a+1}+1 \Rightarrow \\ \\ \Rightarrow 2 \cdot 2^{-a+2}+2 \cdot 1=2^{a-1}+2^{a+1}+1 \Rightarrow \\ \\ \Rightarrow 2 \cdot (2^a)^{-1}\cdot 2^2+2=2^a \cdot 2^{-1}+2^a \cdot 2^1+1 \Rightarrow \\ \\ \Rightarrow 2 \cdot (2^a)^{-1} \cdot 4+2=2^a \cdot \frac{1}{2} +2^a \cdot 2+1 \Rightarrow 8 \cdot (2^a)^{-1}+2=\frac{2^a}{2} +2^a \cdot 2+1\\ \\ Not\v{a}m~2^a=t$

$\displaystyle 8 t^{-1}+2=\frac{t}{2} +2t+1 \Rightarrow \frac{8}{t} +2=\frac{t}{2} +2t+1\Rightarrow  16+4t=t^2+4t^2+2t\Rightarrow \\ \\ \Rightarrow -t^2-4t^2-2t+4t+16=0 \Rightarrow -5t^2+2t+16=0 \Rightarrow \\ \\ \Rightarrow 5t^2-2t-16=0 \\ \\ \Delta=(-2)^2-4 \cdot 5 \cdot (-16)=4+320=324>0$

$\displaystyle t_1=\frac{-(-2)-\sqrt{324} }{2 \cdot 5} =\frac{2-18}{10} =-\frac{16}{10}=-\frac{8}{5} \\ \\ t_2=\frac{-(-2)+\sqrt{324} }{2 \cdot 5} =\frac{2+18}{10} =\frac{20}{10}=2\\ \\ 2^a=-\frac{8}{5} \Rightarrow Nu~sunt~solu\c{t}ii~pentru~a \in \mathbb{R}\\ \\ 2^a=2 \Rightarrow \boxed{a=1}$