[tex] b_{n-1}=x-5 [/tex]
[tex] b_{n}= \frac{2x+1}{2}; [/tex]
[tex] b_{n+1}=3x [/tex]
Tripletele lor: 3(x-5), [tex] \frac{3(2x+1)}{2} [/tex], 9x
Formula pentru progresie geometrica: [tex] b_{n}[/tex]=√[tex] b_{n-1}* b_{n+1} [/tex]
(media geometrica dintre precedent si urmator)
Aplicam: [tex] \frac{3(2x+1)}{2} [/tex]=√3(x-5)*9x |()² ridicam la patrat
[tex] \frac{9(2x+1) ^{2}}{4} [/tex]=3*9x(x-5)
[tex] \frac{9(4x^{2} +8x+1)}{4} [/tex]=3(9x²-45x) |(*4)
9(4x²+8x+1)=12(9x²-45x) | /3
3(4x²+8x+1)=4(9x²-45x)
12x²+24x+3=36x²-180x
-24x²+204x+3=0 |/3 <=> -8x²+68x=0
68x-8x²=0 | :4
17x-2x²=0
x(17-2x)=0
x₁=0, x₂=17/2
Pentru x=0 =>3*(-5)=-15, 3/2, 0 ; nu convine
Pentru x= 17/2 => 21/2, 27, 153/2
