<s>[INST] Sum to infinity An infinite G.P. has a finite sum with initial term $u_1 = 2$. Find the sum to infinity of this G.P. If it is also a Fibonacci sequence
My approach- 


*

*since it is a G.P., $ u_3 = u_2 * u_2/u_1$

*since it is a Fibonacci sequence, $u_3 = u_1+u_2$

*Therefore, $u_2 * u_2/u_1  = u_1+ u_2$


By substituting $u_1 = 2$ and solving, I got $u_2= 1 \pm \sqrt{5}$. 
When $u_2 = 1+\sqrt{5}$ ,


*

*$r = u_2/u_1 = (1+\sqrt{5})/2  \approx 1.618$

*$| r | = | 1.618 |$ = 1.618 > 1 

*therefore the sum to infinity doesn't exist. 


when $u_2 = 1-\sqrt{5}$ ,


*

*$r = u_2/u_1 = (1-\sqrt{5})/2 \approx -0.618$

*$| r | = | -0.618 | = 0.618 < 1$
therefore the sum to infinity exists. 


therefore, a= 2, r = (1-√5)/2
the sum to infinity $\sum_{i=1}^\infty u_i = a / ( 1-r) = -1+\sqrt{5}$. 
Can anyone pls help me if this is true? 
Sorry for my poor typing form. 
 [/INST] Yes, your reasoning looks correct. I get the same answer you do in a slightly different form:
$$\begin{align*}
\sum_{i=1}^\infty u_i &= \frac{a}{1-r}\\
& = \frac{2}{1-\frac{1-\sqrt{5}}{2}}\\
&= \frac{4}{2-(1-\sqrt{5})}\\
&= \frac{4}{1+\sqrt{5}}
\end{align*}$$
(Since $1-\sqrt{5} < 0$, every other term is negative and the overall sum is less than the first term, $a=2$. )
Note that this is answer is equal to $-1+\sqrt{5}$, because:
$$\frac{4}{1+\sqrt{5}} = \frac{4(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})} = \frac{4(1-\sqrt{5})}{1-5} = \frac{4}{-4}(1-\sqrt{5})=-1+\sqrt{5}$$
</s>