[tex]\it \mathbf{12}.\ \ \dfrac{..}{..}\\ \\ \\ \dfrac{b_1}{b_4}=8 \Rightarrow \dfrac{b_1}{b_1q^3}=8 \Rightarrow \dfrac{1}{q^3}=2^3 \Rightarrow \Big(\dfrac{1}{q}\Big)^3=2^3 \Rightarrow \dfrac{1}{q}=2 \Rightarrow q=\dfrac{1}{2}\\ \\ \\ b_2=5 \Rightarrow b_1q=5 \Rightarrow b_1\cdot\dfrac{1}{2}=5 \Rightarrow b_1=10[/tex]
[tex]\it \mathbf{14}.[/tex]
Trei termeni consecutivi ai unei progresii geometrice au
proprietatea că termenul din mijloc este egal cu media geometrică a termenilor vecini.
[tex]\it a)\ \sqrt{\sqrt2\cdot\sqrt5}=\sqrt3 \Rightarrow \sqrt{10}=3\ (F)\\ \\ b)\ \sqrt{\sqrt7\cdot\sqrt9}=\sqrt8 \Rightarrow \sqrt{63}=8\ (F)\\ \\ c)\ \sqrt{\sqrt n\cdot\sqrt{n+2}}=\sqrt{n+1} \Rightarrow \sqrt{n(n+2)}=n+1 \Rightarrow n(n+2)=(n+1)^2\Rightarrow\\ \\ \Rightarrow n^2+2n=n^2+2n+1 \Rightarrow 0=1\ (F)[/tex]
