Q. A geometric series has first term $4$ and common ratio $r$. Where $0 < r < 1$
The first, second and fourth terms of this geometric series form three successive terms of an arithmetic series.
The sum of infinity of the geometric series is.
A. $\frac12 (\sqrt5 - 1)$ B. $2 (3 - \sqrt5)$ C. $2 (1 + \sqrt5)$ D. $2 (3 + \sqrt5)$
Can anyone help me with the answer and explain the method used?
I am really stuck any help would be really appreciated.
Thanks in advance.
Let the geometric series be: $a, ar, ar^2, ar^3$,...
The terms 1, 2 and 4 form an arithmetic series: $a, ar, ar^3$
since the difference between every 2 consecutive terms in arithmetic series is constant:
$ar-a=ar^3-ar$
$ar^3-2ar+a=0$
$a(r-1)(r^2 + r-1)=0$
this leads to 3 possible values of r (assuming $a$ is not zero)
(a) $r=1$. This value leads to a series with infinite sum. Ignore $r=1$.
(b) $r=(-1-\sqrt5)/2$. This is a negative value violating the condition $r>0$.
(c) $r=(-1+\sqrt5)/2$. This value is about .62 and is positive and less than 1, so it satisfies the constraint on r.
now since you know the value of $r$ and you know the sum of an infinite series such as: $a, ar, ar^2,...$ to be:
$S=a/(1-r)$
you can plug in the value of $r$ from (c) above to get:
$S= \frac{a}{1-(\frac{-1+\sqrt5}{2})}$
$S=\frac{2a}{3-\sqrt 5}$
Since a=4; $S=\frac{8}{3-\sqrt 5}$
$S=S \frac{3+\sqrt 5}{3+\sqrt 5}$
$S=\frac{8(3+\sqrt 5)}{9-5}=2(3+\sqrt 5)$
So the correct answer is (D).